yql - Easy Jquery: How to display source -

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i'm sure easy question cannot find example me figure out answer:

i creating query using yql , want replace of values variables. i'm not sure have correct love jquery display effort can test i've created.

this know works:

<script> function handleresponse (json) { var ul = document.getelementsbytagname( 'ul' )[0],     li = null;   ( var = 0; < json.query.count; i++ ) {             li = document.createelement( 'avgrent' );             li.innerhtml = json.results[i];      ul.appendchild( li ); } } </script> <script src="http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3d%22http%3a%2f%2fwww.zillow.com%2fwidgets%2fzestimate%2fzestimatesmallwidget.htm%3fdid%3dzillow-shv-small-iframe-widget%26type%3diframe%26forrent%3dtrue%26address%3d'5894%2bdogwood%2bcir%2b35111'%22%20and%20xpath%3d'%2f%2fdiv%5b%40id%3d%22zestimate-rate-container%22%5d%2fspan%5b2%5d'&diagnostics=true'&callback=handleresponse"></script>

and i'm trying work with:

<script src="http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3d%22http%3a%2f%2fwww.zillow.com%2fwidgets%2fzestimate%2fzestimatesmallwidget.htm%3fdid%3dzillow-shv-small-iframe-widget%26type%3diframe%26forrent%3dtrue%26address%3d'"+ address.replace(/\ /g, '%2b') +"%2b"+postcode+"'%22%20and%20xpath%3d'%2f%2fdiv%5b%40id%3d%22zestimate-rate-container%22%5d%2fspan%5b2%5d'&callback=handleresponse"></script>

instead of actual value (which working) need jquery display src can see if modifications correct. need do?


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