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  • php 报错如下:Notice: Trying to get property of non-object

    参考文档如下解决:

    https://stackoverflow.com/questions/5891911/trying-to-get-property-of-non-object-in

    问题如下:

    This question already has an answer here:

    on Control page:

    <?php
      include 'pages/db.php'; 
      $results = mysql_query("SELECT * FROM sidemenu WHERE `menu_id`='".$menu."' ORDER BY `id` ASC LIMIT 1", $con);
      $sidemenus = mysql_fetch_object($results);
    ?>

    on View Page:

    <?php foreach ($sidemenus as $sidemenu): ?>
      <?php echo $sidemenu->mname."<br />";?>
    <?php endforeach; ?>

    Error is:

    Notice: Trying to get property of non-object in C:wampwwwphonepagesinit.php on line 22

    Can you fix it? I don't have any idea what happened.

    解决如下:

    Check the manual for mysql_fetch_object(). It returns an object, not an array of objects.

    I'm guessing you want something like this

    $results = mysql_query("SELECT * FROM sidemenu WHERE `menu_id`='".$menu."' ORDER BY `id` ASC LIMIT 1", $con);
    
    $sidemenus = array();
    while ($sidemenu = mysql_fetch_object($results)) {
        $sidemenus[] = $sidemenu;
    }

    Might I suggest you have a look at PDO. PDOStatement::fetchAll(PDO::FETCH_OBJ) does what you assumed mysql_fetch_object() to do

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  • 原文地址:https://www.cnblogs.com/feiyun8616/p/6927793.html
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