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Home/ Questions/Q 6919205
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T09:57:37+00:00 2026-05-27T09:57:37+00:00

I have this URL- http://localhost/app_demo/sample.php?jsonRequest={GenInfo:{type:Request,appname:XXX,appversion:1.0.0},searchDish:{userId:295,dishName:,est:Pizza &amp; Wings,location:,type:,priceRange:,deviceos:value,deviceId:<UDID>,deviceType:value,pageNo:1}} when I hit this URL and print

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I have this URL-

http://localhost/app_demo/sample.php?jsonRequest={"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"Pizza &amp; Wings","location":"","type":"","priceRange":"","deviceos":"value","deviceId":"<UDID>","deviceType":"value","pageNo":"1"}}

when I hit this URL and print

print_r($_REQUEST['jsonRequest']);

string print only upto

{"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"2 Pizza

I search the net but did not get the answer.What is solution for this?
please help,
thanks.

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  1. Editorial Team
    Editorial Team
    2026-05-27T09:57:38+00:00Added an answer on May 27, 2026 at 9:57 am

    A query string is normally composed of key/value pairs, the start of a query string is the question mark (?), and then all pairs are separated with an ampersand (&). Having an ampersand in your value is like starting a new parameter.

    However, this is not the right way to do this. You shouldn’t put JSON in the query string.

    If you really must have an ampersand in the query string, use %26 and not &amp. %26 which is the hex value for the ampersand.

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