dongyan1993 2010-04-06 10:48
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如何在jquery中传递参数

I am supposed to submit 2 values called str and name to another page.But only 1value is getting returned in the next page. When I include name and on next page if I say print_r($_POST),then even the first value is not getting printed. I have written a function as follows,which works because there is only one parameter.

function sendValue(str)
{
$.post(
"newsletter/subscribe.php", //Ajax file
{
     sendValue: str 
},
function(data){
        $('#display').html(data.returnValue);
           },
    "json"
);
}

But if I pass 2 values in that function and in $.post, I do sendValue:str,name then I am not getting even 1 value.

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  • dongluoheng3324 2010-04-06 10:56
    关注

    You can post 2 values like this:

    function sendValue(str, name) {
      $.post("newsletter/subscribe.php",
             { 'string': str, 'name' : name },
             function(data){
               $('#display').html(data.returnValue);
             }, 
            "json");
    }
    

    The format for the data argument of $.post() is like this:

    { 'varName' : variable, 'var2Name', variable2, 'var3Name' : variable3 }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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