Uncaught SyntaxError: Unexpected end of input
I have a php nested array stored in a variable $myArray , below is how the array looks like (its not a complete output) after var dumping it to the browser.
<?php var_dump($myArray); ?>
The output:
array (size=4)
'id' => string '162' (length=3)
'content' => string 'Test content' (length=12)
'children' =>
array (size=16)
0 =>
array (size=4)
'id' => string '29208' (length=5)
'content' => string 'Test content 1' (length=14)
'children' =>
array (size=3)
...
1 =>
array (size=4)
'id' => string '29215' (length=5)
'content' => string 'Test content 2' (length=14)
'children' =>
array (size=1)
...
2 =>
array (size=3)
'id' => string '29220' (length=5)
'content' => string 'Test Content 3' (length=14)
Reading the variable array from JavaScript as below:
<script type="text/javascript">
var myVar = JSON.parse('<?php json_encode($myArray) ?>');
</script>
Returns the following error in the console
Uncaught SyntaxError: Unexpected end of input
While debugging the code, I did the following:
Created a new variable and stored some JSON data in it and then JSON parsed it to another variable and then finally consoled the output and it worked fine.
<script type="text/javascript">
var x = '{"id":123,"content":"This is a test content"}';
var myVar = JSON.parse(x);
console.log(myVar);
</script>
The output was an object with those values in the console:
Object
content: "This is a test content"
id: 123
What am I doing wrong?
var myVar = <?php echo json_encode($myArray) ?>; should do it. No ' characters are needed because the JSON object can be read as written, and no parse is necessary because it's outputting directly onto the page instead of giving it a string
You need to echo out the json object.
<?php json_encode($myArray) ?>
to
<?php echo json_encode($myArray) ?>
Here's a little shorthand trick for you explained here
You can simply do <?=$var?> . It's basically shorthand for echo and only works if the shorthand tag <? is enabled.
So the answer to your question (if shorthand open tags are enabled) you can use this
var myVar = <?=json_encode($myArray)?>; Which is equivalent to what @Dar gave you above but less ugly.
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