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  1. PO
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    <p>I tried your function and it worked fine for me when I left out this line: <code>contentType: "application/json",</code> and when I added it in, I started getting a failed response of "0".</p> <p>In the <a href="http://api.jquery.com/jQuery.ajax/" rel="nofollow">manual</a>, it says to use this content type for most cases: <code>'application/x-www-form-urlencoded; charset=UTF-8'</code> (which is the default). So unless you have a good reason, I think you should remove that line.</p> <p>In the first function, it doesn't exist and you turned off 'json' dataType, so you are probably alerting a really nice looking string that <em>looks</em> like the data you want. If you make sure that you have <code>dataType: 'json'</code> and remove <code>contentType: application/json'</code>, you should be good to go.</p> <p><strong>EDIT</strong>: This is exactly what I used, and it worked great. Not sure what else to suggest. What is <code>datastr</code> in your js?</p> <pre><code>$.ajax({ url: 'dbtryout2_2.php', type: 'post', async: true, cache: false, dataType: 'json', success: function( response ) { // response is returned as json object: ["album1","album2","album3","album4","album5"] //console.log(response); if ( response ) { $.each(response, function (i, v) { alert(i + "--" + v); }); } }, error: function( xhr ) { console.log(xhr.responseText); } }); </code></pre> <p><strong>Note</strong>: If you have specified "dataType:'json'", you don't have to $.parseJSON, because the response is already returned as JSON.</p>
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    1. POhow to display elements of json array returned from php within a loop in ajax(jquery)
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    1. PTAnswer
    1. USJake
    1. USJake
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    1. CO@ jake still its not working.see the code above . i have clarified my problem again.
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      1. USNitish kumar thakur
    2. COwell finally the above code is working.U know actually the problem was in the data that i was returning from php. I was echoing two things from php one was a "message" that i had written initially for checking whether php code is running or not and the other was the "array" that actually i wanted to return.But i was not at all concentrating on the string that was getting returned with the array.
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      1. USNitish kumar thakur
    3. COwell thanks a lot to all of you for your suggestions
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      1. USNitish kumar thakur
 

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