dto52236 2017-07-30 19:31
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将MySQL转换为MySQLi问题[重复]

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This the error

This is error line 125. Actually, I am new to MySQLi, so I am not able to understand how to convert MySQL to MySQLi.

This is my code:

<?php 

$query = mysql_query("select * from upload ORDER BY id DESC") or die(mysql_error());

while ($row = mysql_fetch_array($query)) {
    $id = $row['id'];
    $name = $row['name'];
    $date=$row['date'];
}
</div>
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1条回答 默认 最新

  • duan6301 2017-07-30 19:44
    关注

    You need to pass in the connection as a parameter to mysqli_query():

    $connection = mysqli_connect(
        "your_host",
        "your_user",
        "your_password",
        "your_db"
    );
    
    $result = mysqli_query(
        $connection, 
        "select * from upload ORDER BY id DESC"
    );
    
    if (false === $result) {
        die(mysqli_error($connection));
    }
    
    while ($row = mysqli_fetch_array($result)) {
        $id = $row['id'];
        $name = $row['name'];
        $date = $row['date'];
    }
    

    For reference, see:

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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