在foreach中使用MySQL请求时,首先选择是好的,但不是

使用此代码:

foreach ($content as $value) {
    $data=$value[0];
    echo $data; 
    $req="SELECT * FROM TABLE WHERE data='$data'";
    $result=mysql_query($req) or die ('Erreur :'.mysql_error());
    if (mysql_num_rows($result)){
        echo '  ENTRY EXISTS';
    }
    else {
        echo '  ENTRY DOES NOT EXIST';
    }   
}

对于第一个,$value它找到一个条目,这是正确的.对于下一个它没有,但它应该.怎么解决这个问题？

更新代码

使用此代码:

        $found_list = array();

        $fetch_list = array();
        foreach($content as $value){
            $fetch_list[] = "'" . mysql_real_escape_string($value[0]) . "'";
        }

        if( empty($fetch_list) ){
            echo '
No data to fetch
';
        }else{

            $sql = 'SELECT DISTINCT inst_name
                FROM INSTITUTS
                WHERE inst_name IN (' . implode(', ', $fetch_list) . ')';
            $res = mysql_query($sql)
                or die ('Error: ' . mysql_error());
            while( $row = mysql_fetch_assoc($res) ){
                $found_list[] = $row['inst_name'];
            }
            var_dump($found_list);
        }

        foreach($content as $value){
            echo '
';
            echo $value[0] . ' ';
            if( in_array($value[0], $found_list) ){
                echo "ENTRY EXISTS\n 
";
            }else{
                echo "ENTRY DOES NOT EXIST\n 
";
            }
        }

结果是:

 array(3) { [0]=> string(13) "AixEnProvence" [1]=> string(19) "AixEnProvenceAnnexe" [2]=> string(7) "Acheres" } 
acheres ENTRY DOES NOT EXIST 
AixEnProvence ENTRY EXISTS 
aixenprovenceannexe ENTRY DOES NOT EXIST 
instituttest ENTRY DOES NOT EXIST

Álvaro Gonzá.. 5

没有理由用几乎相同的查询来淹没MySQL服务器.看看IN表达式:

SELECT foo, bar
FROM table
WHERE data IN ('a', 'b', 'c');

我还建议你谷歌搜索SQL注入和XSS攻击.

编辑:这里有一些代码解决了最新评论中描述的问题:

No data to fetch
';
}else{
    $sql = 'SELECT DISTINCT data
        FROM table
        WHERE data IN (' . implode(', ', $fetch_list) . ')';
    $res = mysql_query($sql)
        or die ('Error: ' . mysql_error());
    while( $row = mysql_fetch_assoc($res) ){
        $found_list[] = $row['data'];
    }
}

foreach($content as $value){
    echo $value[0] . ' ';
    if( in_array($value[0], $found_list) ){
        echo "ENTRY EXISTS\n";
    }else{
        echo "ENTRY DOES NOT EXIST\n";
    }
}

?>

回答更新的问题:

PHP 比较运算符区分大小写:

您可以使用strtolower()在比较之前标准化值.

没有理由用几乎相同的查询来淹没MySQL服务器.看看IN表达式:

SELECT foo, bar
FROM table
WHERE data IN ('a', 'b', 'c');

我还建议你谷歌搜索SQL注入和XSS攻击.

编辑:这里有一些代码解决了最新评论中描述的问题:

No data to fetch
';
}else{
    $sql = 'SELECT DISTINCT data
        FROM table
        WHERE data IN (' . implode(', ', $fetch_list) . ')';
    $res = mysql_query($sql)
        or die ('Error: ' . mysql_error());
    while( $row = mysql_fetch_assoc($res) ){
        $found_list[] = $row['data'];
    }
}

foreach($content as $value){
    echo $value[0] . ' ';
    if( in_array($value[0], $found_list) ){
        echo "ENTRY EXISTS\n";
    }else{
        echo "ENTRY DOES NOT EXIST\n";
    }
}

?>

回答更新的问题:

PHP 比较运算符区分大小写:

您可以使用strtolower()在比较之前标准化值.