reference error in javascript array

Question

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].

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. I have placed the php code before tag in the page and the javascript code is in the body at the last position.

$sql= mysql_query( "select * from questions where Q_ID like '{$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. But everytime I execute this code it shows reference error in firebug console window.

ReferenceError: CSE6014 is not defined record[0]=CSE6014;/* Showing refernce error (ReferenceError: CSE6014 is not defi...

I don't know what I am doing wrong. Please help me. Thanks in advance.

Can you please tell me how to use statement like this if(record.length< <?php echo $NoOfQuestion-1; ?>) where $NoOfQuestion is an integer. — vishnu priya
– vishnu priya, Commented Jul 3, 2015 at 9:11
I am retrieving its value from database. Consider it can be any integer value (maximum 100). — vishnu priya
– vishnu priya, Commented Jul 3, 2015 at 9:15
You can simply make a use of another variable as var myvar = <?php echo $NoOfQuestion; ?>-1 and if(record.length< myvar; ?>) and be sure that the value within your php code must be int and not a string — Narendrasingh Sisodia
– Narendrasingh Sisodia, Commented Jul 3, 2015 at 10:08

Tushar · Accepted Answer · 2015-07-03 07:18:33Z

3

As $Q_ID and $corr are strings, you need to surround string with quotes. You can use either single quote ' or double quote ".

Use following code(Notice the quotes around the PHP tags):

record[0] = '<?php echo $Q_ID ?>';
correct[0] = '<?php echo $corr ?>';

answered Jul 3, 2015 at 7:18

Tushar

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6 Comments

vishnu priya Over a year ago

Sorry sir but its not working in this case. "$NoOfQuestion" is a number not string.

Tushar Over a year ago

@vishnupriya Try if(record.length< parseInt('<?php echo $NoOfQuestion-1; ?>')

vishnu priya Over a year ago

It is giving this error in firebug console SyntaxError: unterminated string literal if(record.length< parseInt('<br />

Tushar Over a year ago

@vishnupriya if(record.length< parseInt('<?php echo $NoOfQuestion-1; ?>'))

vishnu priya Over a year ago

Sir it is giving unterminated string literal syntax error

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