当echo $ _GET [“" jsoncallback“]时发生错误 [英] Error when echo $_GET["jsoncallback"]

问题描述

<?php    
$query = mysql_query("Select id, name From table");
        while($row = mysql_fetch_array($query)) {
            echo $_GET["jsoncallback"] . '(<option value='.$row['id'].'>'.$row['name'].'</option>)';
        }
?>

当我回显结果时，这是错误，如何解决?

When I echo result, it is error, how to fix it?

推荐答案

由于您未完全编写要执行的操作，因此我猜测您正在尝试返回JS的HTML选项列表回调函数将放置在您的文档中.

Since you didn't write exactly what you're trying to do, i'm guessing you're trying to return a list of HTML options that a JS callback function will place in your document.

尝试一下:

<?php    
$options = '';
$query = mysql_query("Select id, name From table");
while ($row = mysql_fetch_array($query)) {
  $options .= '<option value="'.$row['id'].'">'.$row['name'].'</option>' . "\n";
}
echo $_GET["jsoncallback"] . "('" . $options . "');";
?>

这将首先将所有选项创建为字符串，然后再构建回调.

This will first create all the options as a string, and only then build the callback.

这篇关于当echo $ _GET [“" jsoncallback“]时发生错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！