Question

我试图复制How to Check if value exists in a MySQL database并制作我自己的，但由于某种原因它不会工作......

这就是我得到的：

<?php
$host = '127.0.0.1';
$username = 'root';
$password = '';
$dbname = 'multiplayer';
$con=mysqli_connect($host, $username, $password, $dbname);

$check_player_ip=mysqli_query($con, 'SELECT `player_ip` FROM `playerdata` WHERE username = "remco" AND active = 0');
    if (mysqli_num_rows($check_player_ip) == 0) {
      //didnt find anything
    } else{
      //found something
    }
?>

我收到此错误： 解析错误：语法错误，意外＆＃39; $ check_player_ip＆＃39;第1行的C：\ xampp \ htdocs \ test.php中的（T_VARIABLE）

解决方案

如果收到T_VARIABLE错误，请检查此规则之前的变量。你可能会忘记＆＃39 ;;＆＃39;的xD

感谢所有支持！

Answer 1

<击> 您正在混合MySQL API，它们不会混合在一起。 mysql_num_rows

<击>

使用mysqli_num_rows()

另外，请确保您的数据库连接也是mysqli_*而不是mysql_*

<击>

编辑完问题后，

编辑。

您需要将DB连接传递给您的查询：

$check_player_ip=mysqli_query($con,'SELECT...`

$con是您的数据库连接。相应地改变。

加WHERE username = remco - 单词remco需要用引号括起来，它是一个字符串而不是int。

WHERE username = 'remco'

<强> 旁注：

您现有的代码向SQL injection开放使用prepared statements或PDO with prepared statements。

编辑2：

尝试反转引号：

$check_player_ip=mysqli_query($con, "SELECT `player_ip` FROM `playerdata` WHERE username = 'remco' AND active = 0");
    if (mysqli_num_rows($check_player_ip) == 0) {
      //didnt find anything
    }

Answer 2

你应该执行那个mysqli查询...

while($rows = mysql_arrayAssoc($ursql)){
    $data[]=$rows;
}
if($something== $data['attribute']) //attribute(id,name...)
    echo "ok some data is in"
else 
   echo "no matching data"

我希望它可以帮到你:)。

Answer 3

试试这个：

<?php 
$host = '127.0.0.1';
$username = 'root';
$password = '';
$dbname = 'multiplayer';

$con = new mysqli( $host, $username, $password, $dbname );

/* Check Connection */
    if ( $con->connect_error ) {
        printf( "Connect failed: %s\n", $con->connect_error );
        exit();
    }

/* Query - Returns a resultset */
    if ( $result = $con->query('SELECT `player_ip` FROM `playerdata` WHERE username = "remco" AND active = 0 ') ) {

    if ( $result->num_rows <= 0 ) {
      //didnt find anything
          printf("No player");
    } else {
      //found something
          printf("Select returned %d rows.\n", $result->num_rows);
    }

    /* free result set */
        $result->close();
}

/* close connection */
    $con->close();
?>

PHP检查MySQL中是否存在值

3 个答案: