MartinReviewHandler::SaveReview()   B
last analyzed

Complexity

Conditions 6
Paths 9

Size

Total Lines 22

Duplication

Lines 0
Ratio 0 %

Importance

Changes 0
Metric Value
cc 6
nc 9
nop 1
dl 0
loc 22
rs 8.9457
c 0
b 0
f 0
1
<?php
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/**
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 *
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 * @access    public
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 * @return void
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 * @copyright 1997-2010 The Lap Group
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 * @author    Martin <[email protected]>
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 * @created   time :2010-07-22 16:27:07
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 * */
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class MartinReview extends XoopsObject
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{
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}
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/**
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 *
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 * @access    public
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 * @return void
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 * @copyright 1997-2010 The Lap Group
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 * @author    Martin <[email protected]>
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 * @created   time :2010-07-22 16:27:25
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 * */
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class MartinReviewHandler extends XoopsObjectHandler
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{
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    /**
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     * @create    cart object
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     * @license   http://www.blags.org/
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     * @created   :2010年07月04日 12时59分
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     * @copyright 1997-2010 The Martin Group
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     * @author    Martin <[email protected]>
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     * */
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    public function &create()
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    {
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        $obj =& new MartinReview;
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        return $obj;
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    }
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    /**
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     * insert review for hotel
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     * @access    public
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     * @param $Data
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     * @copyright 1997-2010 The Lap Group
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     * @author    Martin <[email protected]>
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     * @created   time :2010-07-22 16:28:18
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     */
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    public function SaveReview($Data)
0 ignored issues
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Documentation introduced by
The return type could not be reliably inferred; please add a @return annotation.

Our type inference engine in quite powerful, but sometimes the code does not provide enough clues to go by. In these cases we request you to add a @return annotation as described here.

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48
    {
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        global $xoopsDB;
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        if (!$Data) {
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            return $Data;
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        }
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        $table = $xoopsDB->prefix('martin_user_review');
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        $exist = $this->CheckReviewExist($Data['hotel_id'], $Data['uid']);
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        if (is_array($Data)) {
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            foreach ($Data as $key => $value) {
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                $v .= $prefix . $value;
0 ignored issues
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Bug introduced by
The variable $v does not seem to be defined for all execution paths leading up to this point.

If you define a variable conditionally, it can happen that it is not defined for all execution paths.

Let’s take a look at an example:

function myFunction($a) {
    switch ($a) {
        case 'foo':
            $x = 1;
            break;

        case 'bar':
            $x = 2;
            break;
    }

    // $x is potentially undefined here.
    echo $x;
}

In the above example, the variable $x is defined if you pass “foo” or “bar” as argument for $a. However, since the switch statement has no default case statement, if you pass any other value, the variable $x would be undefined.

Available Fixes

  1. Check for existence of the variable explicitly:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        if (isset($x)) { // Make sure it's always set.
            echo $x;
        }
    }
    
  2. Define a default value for the variable:

    function myFunction($a) {
        $x = ''; // Set a default which gets overridden for certain paths.
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        echo $x;
    }
    
  3. Add a value for the missing path:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
    
            // We add support for the missing case.
            default:
                $x = '';
                break;
        }
    
        echo $x;
    }
    
Loading history...
Bug introduced by
The variable $prefix does not seem to be defined for all execution paths leading up to this point.

If you define a variable conditionally, it can happen that it is not defined for all execution paths.

Let’s take a look at an example:

function myFunction($a) {
    switch ($a) {
        case 'foo':
            $x = 1;
            break;

        case 'bar':
            $x = 2;
            break;
    }

    // $x is potentially undefined here.
    echo $x;
}

In the above example, the variable $x is defined if you pass “foo” or “bar” as argument for $a. However, since the switch statement has no default case statement, if you pass any other value, the variable $x would be undefined.

Available Fixes

  1. Check for existence of the variable explicitly:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        if (isset($x)) { // Make sure it's always set.
            echo $x;
        }
    }
    
  2. Define a default value for the variable:

    function myFunction($a) {
        $x = ''; // Set a default which gets overridden for certain paths.
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        echo $x;
    }
    
  3. Add a value for the missing path:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
    
            // We add support for the missing case.
            default:
                $x = '';
                break;
        }
    
        echo $x;
    }
    
Loading history...
58
                $k .= $prefix . $key;
0 ignored issues
show
Bug introduced by
The variable $k does not seem to be defined for all execution paths leading up to this point.

If you define a variable conditionally, it can happen that it is not defined for all execution paths.

Let’s take a look at an example:

function myFunction($a) {
    switch ($a) {
        case 'foo':
            $x = 1;
            break;

        case 'bar':
            $x = 2;
            break;
    }

    // $x is potentially undefined here.
    echo $x;
}

In the above example, the variable $x is defined if you pass “foo” or “bar” as argument for $a. However, since the switch statement has no default case statement, if you pass any other value, the variable $x would be undefined.

Available Fixes

  1. Check for existence of the variable explicitly:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        if (isset($x)) { // Make sure it's always set.
            echo $x;
        }
    }
    
  2. Define a default value for the variable:

    function myFunction($a) {
        $x = ''; // Set a default which gets overridden for certain paths.
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        echo $x;
    }
    
  3. Add a value for the missing path:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
    
            // We add support for the missing case.
            default:
                $x = '';
                break;
        }
    
        echo $x;
    }
    
Loading history...
59
                $updateStr .= $prefix . "$key = $value";
0 ignored issues
show
Bug introduced by
The variable $updateStr does not seem to be defined for all execution paths leading up to this point.

If you define a variable conditionally, it can happen that it is not defined for all execution paths.

Let’s take a look at an example:

function myFunction($a) {
    switch ($a) {
        case 'foo':
            $x = 1;
            break;

        case 'bar':
            $x = 2;
            break;
    }

    // $x is potentially undefined here.
    echo $x;
}

In the above example, the variable $x is defined if you pass “foo” or “bar” as argument for $a. However, since the switch statement has no default case statement, if you pass any other value, the variable $x would be undefined.

Available Fixes

  1. Check for existence of the variable explicitly:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        if (isset($x)) { // Make sure it's always set.
            echo $x;
        }
    }
    
  2. Define a default value for the variable:

    function myFunction($a) {
        $x = ''; // Set a default which gets overridden for certain paths.
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
        }
    
        echo $x;
    }
    
  3. Add a value for the missing path:

    function myFunction($a) {
        switch ($a) {
            case 'foo':
                $x = 1;
                break;
    
            case 'bar':
                $x = 2;
                break;
    
            // We add support for the missing case.
            default:
                $x = '';
                break;
        }
    
        echo $x;
    }
    
Loading history...
60
                $prefix = ',';
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            }
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        }
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        $sql = $exist ? "UPDATE $table SET %s WHERE hotel_id = {$Data['hotel_id']} AND uid = {$Data['uid']}" : "INSERT INTO $table (%s) VALUES (%s)";
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        $sql = $exist ? sprintf($sql, $updateStr) : sprintf($sql, $k, $v);
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        //echo $sql;
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        return $xoopsDB->query($sql);
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    }
69
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    /**
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     *
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     * @access    public
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     * @param $hotel_id
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     * @param $uid
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     * @return bool
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     * @copyright 1997-2010 The Lap Group
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     * @author    Martin <[email protected]>
78
     * @created   time :2010-07-22 16:43:23
79
     */
80 View Code Duplication
    public function CheckReviewExist($hotel_id, $uid)
0 ignored issues
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Duplication introduced by
This method seems to be duplicated in your project.

Duplicated code is one of the most pungent code smells. If you need to duplicate the same code in three or more different places, we strongly encourage you to look into extracting the code into a single class or operation.

You can also find more detailed suggestions in the “Code” section of your repository.

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81
    {
82
        global $xoopsDB;
83
        $sql = "SELECT * FROM {$xoopsDB->prefix('martin_user_review')} WHERE hotel_id = $hotel_id AND uid = $uid ";
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        return is_array($xoopsDB->fetchArray($xoopsDB->query($sql)));
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    }
87
88
    /**
89
     *
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     * @access    public
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     * @param $hotel_id
92
     * @copyright 1997-2010 The Lap Group
93
     * @author    Martin <[email protected]>
94
     * @created   time :2010-07-22 16:49:41
95
     */
96
    public function GetReview($hotel_id)
0 ignored issues
show
Documentation introduced by
The return type could not be reliably inferred; please add a @return annotation.

Our type inference engine in quite powerful, but sometimes the code does not provide enough clues to go by. In these cases we request you to add a @return annotation as described here.

Loading history...
97
    {
98
        global $xoopsDB, $xoopsUser;
99
        $uid = $xoopsUser->uid();
100
        $sql = "SELECT * FROM {$xoopsDB->prefix('martin_user_review')} WHERE hotel_id = $hotel_id AND uid = $uid ";
101
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        return $xoopsDB->fetchArray($xoopsDB->query($sql));
103
    }
104
105
    /**
106
     * get hotel review
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     * @access    public
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     * @param $hotel_id
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     * @copyright 1997-2010 The Lap Group
110
     * @author    Martin <[email protected]>
111
     * @created   time :2010-07-22 17:03:32
112
     */
113
    public function GetHotelReview($hotel_id)
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    {
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        global $xoopsDB;
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        $table                  = $xoopsDB->prefix('martin_user_review');
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        $sql                    = "SELECT avg(review_type_avg) as review_type_avg , count(uid) as count FROM $table WHERE hotel_id = $hotel_id GROUP BY hotel_id ";
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        $row                    = ($xoopsDB->fetchArray($xoopsDB->query($sql)));
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        $row['review_type_avg'] = round($row['review_type_avg'], 2);
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        $row['score']           = (int)(($row['review_type_avg'] / 5) * 100);
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        //var_dump($row);
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        return $row;
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    }
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}
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