[英]reference error in javascript array
I need to store the value of php variable $Q_ID in javascript array at index 0. Here is my code for storing it in javascript array record[0]. 我需要将php变量$ Q_ID的值存储在索引0的javascript数组中。这是我将其存储在javascript数组记录[0]中的代码。
var record= [];
var choice= [];
var correct=[];
record[0]=<?php echo $Q_ID ?>;/* Showing refernce error (ReferenceError: CSE6014 is not defined record[0]=CSE6014) */;
correct[0]=<?php echo $corr ?>;
And this is the php code to get the value of Q_ID. 这是获取Q_ID值的php代码。 I have placed the php code before tag in the page and the javascript code is in the body at the last position.
我已经在页面中的标记之前放置了php代码,并且javascript代码位于最后位置的正文中。
$sql= mysql_query( "select * from questions where Q_ID like '{$code}%' order by RAND() limit 1" ) or die(mysql_error());
$rows = mysql_fetch_array($sql);
$Q_ID = $rows['Q_ID'];
$question= $rows['Question'];
$opt1=$rows['Option_1'];
$opt2=$rows['Option_2'];
I have placed the php code before tag in the page and the javascript code is in the body at the last position. 我已经在页面中的标记之前放置了php代码,并且javascript代码位于最后位置的正文中。 But everytime I execute this code it shows reference error in firebug console window.
但每次执行此代码时,它都会在firebug控制台窗口中显示引用错误。
ReferenceError: CSE6014 is not defined record[0]=CSE6014;/* Showing refernce error (ReferenceError: CSE6014 is not defi... ReferenceError:CSE6014未定义record [0] = CSE6014; / *显示引用错误(ReferenceError:CSE6014不是defi ...
I don't know what I am doing wrong. 我不知道我做错了什么。 Please help me.
请帮我。 Thanks in advance.
提前致谢。
As $Q_ID
and $corr
are strings
, you need to surround string
with quotes
. 由于
$Q_ID
和$corr
是strings
,您需要用quotes
string
。 You can use either single quote '
or double quote "
. 您可以使用单引号
'
或双引号"
。
Use following code(Notice the quotes around the PHP tags): 使用以下代码(注意PHP标记周围的引号):
record[0] = '<?php echo $Q_ID ?>';
correct[0] = '<?php echo $corr ?>';
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