Passing value in HTML URL link

Question

$.post("/diabetes/ropimages/getcount.php",{pid:$("#patient_id").val()} ,function(data1){
  //alert(data1);
  var count = data1;
  var pid = $("#patient_id").val();
  var rid;

  for( var i = 1 ; i <= count ; i++) {
    var link ='<img src="/diabetes/ropimages/thumbpicdisplay.php?pid=+pid+&rid=1" />';
    $("#content1").empty().html(link);
  }
});

I am trying to pass pid value in url ..but its taking directly as +pid+ as value ..how do i give it the value of pid.

And how do i print 3 images in a div? like the one in code

Daniel Vassallo · Accepted Answer · 2010-03-15 21:59:06Z

3

You simply need to terminate the string after ?pid= and then use the concatenation operator (+) to "insert" the pid variable in the appropriate location:

'<img src="/diabetes/ropimages/thumbpicdisplay.php?pid=' + pid + '&rid=1" />'

As for attaching the 3 images to the div, you might have more luck doing the following:

var link = '';

for(var i = 1; i <= count; i++) {
   link += '<img src="...thumbpicdisplay.php?pid=' + pid + '&rid=1" />';
}

$("#content1").empty().html(link);

edited Mar 15, 2010 at 21:59

answered Mar 15, 2010 at 8:04

Daniel Vassallo

346k72 gold badges514 silver badges447 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Hacker Over a year ago

hey i wrote code like this... for 1st loop it works then it stops working can u just tell me where i am going wrong

Hacker Over a year ago

$.getJSON("/diabetes/ropimages/getcount.php",{pid:$("#patient_id").val()} ,function(data1){ var i; var count = data1.count; var start = data1.start; var pid = $("#patient_id").val(); for(i = start ; i <= count ; i++) { var link ='<a href="/diabetes/ropimages/origpicdisplay.php?pid='+pid+'&rid='+i+'"><img src="/diabetes/ropimages/thumbpicdisplay.php?pid='+pid+'&rid='+i+'"/><a>'; alert(link); if(i > start) { $("#content1").append(link); } else{ $("#content1").empty().html(link); } } },"Json");

Daniel Vassallo Over a year ago

@pradeep: What value of count are you getting?

Hacker Over a year ago

i am getting start value as 6 and count as 3 .when i try to add those 2 like count = count+start; and print it ..it gives like 36 instead of 9 . i think its taken it as string ..how to solve this prob?

Daniel Vassallo Over a year ago

you can use parseInt(count) + parseInt(start) to avoid that problem.

Collectives™ on Stack Overflow

Passing value in HTML URL link

1 Answer 1

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related