dqrm8199 2012-02-25 17:07
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用于MySQL SELECT的PHP函数[重复]

Possible Duplicate:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

    function arr($string)
    {
        $tank5 = "SELECT url FROM `db`.`tank` WHERE url LIKE {$string}";
        $result5 = mysql_query($tank5);
            if (!$result5) {
    die('Invalid query: ' . mysql_error());
}

        $parts2 = array();
        while ($p = mysql_fetch_array($result5)) {
            $parts2[] = $p['url'];
        }

        // return the array created.
        return $parts2; 
    }

    $array5 = arr('lol.com');       
    print_r($array5);

Why don't the above code work. The error I'm getting is
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /usr/local/...
so I guess it's not database connection problem. Any idea?

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2条回答 默认 最新

  • doujiaoben28596 2012-02-25 17:09
    关注

    Most likely, you're using a LIKE clause but not passing wildcards with $string. Therefore, it would only find exact matches.

    Instead surround $string with % and single-quotes, since you did not quote the input value (assuming the function's input parameter doesn't already have quotes, which seems unlikely).

    $tank5 = "SELECT url FROM `db`.`tank` WHERE url LIKE '%$string%'";
    

    We assume that prior to calling this function, you have already escaped $string against SQL injection like:

    $string = 'lol.com';
    $string = mysql_real_escape_string($string);
    $arr = arr($string);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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