javascript - send string with two parameters - Stack Overflow

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I have this code, but this line has some problem.

 var dataString = 'name='+name&'id='+id;

what is sent (firebug):

'id ' id
'name' name

The line above works correctly if i do: var dataString = 'name='+name; However, i need to pass two parameters. What is the correct way to do that?

code

 <script type="text/javascript">
    $(function () {
        $(".vote").click(function () {
            var id = $(this).attr("id");
            var name = $(this).attr("name");
            var dataString = 'name='+name&'id='+id;

            if (name == 'up') {
                $.ajax({
                    type: "POST",
                    url: "url.php",
                    data: dataString,
                    cache: false,
                    success: function (html) {

                    }
                });
            return false;
        });
    });
    </script>

I have this code, but this line has some problem.

 var dataString = 'name='+name&'id='+id;

what is sent (firebug):

'id ' id
'name' name

The line above works correctly if i do: var dataString = 'name='+name; However, i need to pass two parameters. What is the correct way to do that?

code

 <script type="text/javascript">
    $(function () {
        $(".vote").click(function () {
            var id = $(this).attr("id");
            var name = $(this).attr("name");
            var dataString = 'name='+name&'id='+id;

            if (name == 'up') {
                $.ajax({
                    type: "POST",
                    url: "url.php",
                    data: dataString,
                    cache: false,
                    success: function (html) {

                    }
                });
            return false;
        });
    });
    </script>

• javascript
• jquery
• parameters

Share Improve this question Follow asked Dec 26, 2011 at 19:14 daniel__daniel__ 11.9k1616 gold badges6767 silver badges9191 bronze badges 2

• & goes in the string.'name='+name&'id='+id; should really be 'name='+name+'&id='+id; – nico Commented Dec 26, 2011 at 19:19
• BTW, in an event handler $(this).attr('id') === this.id. – Alnitak Commented Dec 26, 2011 at 19:22

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 6

You should do:

 var dataString = { name: name, id: id}

instead of

 var dataString = 'name='+name&'id='+id;

So that you are sure that the supplied values are correctly URI encoded.

Try this:

var dataString = 'name='+name+'&id='+id;

Instead of

var dataString = 'name='+name&'id='+id;

The & should be inside '', and you need extra + to concat "name" variable and '&id=' string. So this should work.

UPD:

You can also do:

var dataString = { name: name, id: id }

I have this code, but this line has some problem.

 var dataString = 'name='+name&'id='+id;

what is sent (firebug):

'id ' id
'name' name

The line above works correctly if i do: var dataString = 'name='+name; However, i need to pass two parameters. What is the correct way to do that?

code

 <script type="text/javascript">
    $(function () {
        $(".vote").click(function () {
            var id = $(this).attr("id");
            var name = $(this).attr("name");
            var dataString = 'name='+name&'id='+id;

            if (name == 'up') {
                $.ajax({
                    type: "POST",
                    url: "url.php",
                    data: dataString,
                    cache: false,
                    success: function (html) {

                    }
                });
            return false;
        });
    });
    </script>

I have this code, but this line has some problem.

 var dataString = 'name='+name&'id='+id;

what is sent (firebug):

'id ' id
'name' name

The line above works correctly if i do: var dataString = 'name='+name; However, i need to pass two parameters. What is the correct way to do that?

code

 <script type="text/javascript">
    $(function () {
        $(".vote").click(function () {
            var id = $(this).attr("id");
            var name = $(this).attr("name");
            var dataString = 'name='+name&'id='+id;

            if (name == 'up') {
                $.ajax({
                    type: "POST",
                    url: "url.php",
                    data: dataString,
                    cache: false,
                    success: function (html) {

                    }
                });
            return false;
        });
    });
    </script>

• javascript
• jquery
• parameters

Share Improve this question Follow asked Dec 26, 2011 at 19:14 daniel__daniel__ 11.9k1616 gold badges6767 silver badges9191 bronze badges 2

• & goes in the string.'name='+name&'id='+id; should really be 'name='+name+'&id='+id; – nico Commented Dec 26, 2011 at 19:19
• BTW, in an event handler $(this).attr('id') === this.id. – Alnitak Commented Dec 26, 2011 at 19:22

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 6

You should do:

 var dataString = { name: name, id: id}

instead of

 var dataString = 'name='+name&'id='+id;

So that you are sure that the supplied values are correctly URI encoded.

Try this:

var dataString = 'name='+name+'&id='+id;

Instead of

var dataString = 'name='+name&'id='+id;

The & should be inside '', and you need extra + to concat "name" variable and '&id=' string. So this should work.

UPD:

You can also do:

var dataString = { name: name, id: id }

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