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我有这个 PHP 函数，它将 url 插入 MySQL，但只有那些在执行时才存在的。我有一个问题，我的 if 条件被简单地忽略了，所以一切都进入数据库，忽略了条件。代码在这里：

function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT COUNT(*) FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if($row['COUNT(*)'] < 1);
    {
       echo "<font color='red'>".$row['COUNT(*)']."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}

我尝试了 MySQL 计数和 PHP mysqli_num_rows，但仍然相同。无论 if 语句中的条件是什么，它都会忽略它。请帮忙...

4

3 回答 3

3 
function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT COUNT(*) as CNT FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if($row['CNT'] < 1) {
       echo "<font color='red'>".$row['CNT']."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}

或者

function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT link FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if(mysqli_num_rows($resultcheck)==0) {
       echo "<font color='red'>".mysqli_num_rows($resultcheck)."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}
于 2013-11-14T11:35:04.220 回答
2

试试这个if($row['COUNT(*)'] < 1){，而不是if($row['COUNT(*)'] < 1);{

    function storeLink($url,$gathered_from) {

        global $conn;   
        $querycheck = "SELECT COUNT(*) as TotalRow FROM test WHERE link = '$url'";
        $resultcheck = mysqli_query($conn, $querycheck);
        $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

        if($row['TotalRow'] == 0)
        {
           echo "<font color='red'>".$row['TotalRow']."</font>";
           $url = mysqli_real_escape_string($conn,$url);
           $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
           $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
           mysqli_query($conn,$query) or die('Error, insert query failed');
        }
    }
于 2013-11-14T11:34:14.797 回答
1 
INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')
WHERE NOT EXISTS (
   SELECT COUNT(*) as TotalRow FROM test WHERE link = '$url'
) LIMIT 1;

如您所见，您只需 1 个查询即可完成全部工作

于 2013-11-14T11:43:31.423 回答