似乎无法通过jquery将多个变量传递给mysql

Question

I have seen several examples and can't seem to get the hang of passing more than one variable to mysql using jquery. Here is my situation:

I have a page with 2 cascading drop downs,( they work great using jquery to update second drop down based on the first drop down.)

when the first drop down is selected jquery updates the second drop down AND passes the customer id to a php script that creates a new record in the tblinvoice table (this also works great no problems.)

when the second drop down is selected I need to pass that value along with the invoice number to my php script so I can update the record with the instid.(this is the part that don't work)

If I only pass the instid and manually put the invoice number in the where clause of the query all works fine. If I omit the where clause all records are updated as expected. I need to know what I am doing wrong or what is missing.
I will try to post the code here

jquery code

$(document).ready(function() {

    $("select#cust").change(function() {
        var cust_id = $("select#cust option:selected").attr(
            'value');
        var test = $("#test").val();
        var din = $("#idate").val();
        $("#inst").html("");

        if (cust_id.length > 0) {
            $.ajax({
                type: "POST",
                url: "fetch_inst.php",
                data: "cust_id=" + cust_id,
                cache: false,
                beforeSend: function() {
                    $('#inst').html(
                        '<img src="loader.gif" alt="" width="24" height="24">'
                    );
                },
                success: function(html) {
                    $("#inst").html(html);
                }
            });

            if (test == 0) {
                $.ajax({
                    type: "POST",
                    url: "wo_start.php",
                    data: "cust_id=" + cust_id,
                    cache: false,
                    beforeSend: function() {

                    },
                    success: function(html) {
                        $("#invoice").html(html);
                        $("#test").val(1);
                        var inum = $("#inv").val();
                        $("#invnum").val(din +
                            "-" + inum);
                    }
                });
            }
        }
    });


    $("select#inst").change(function() {
        var inst_id = $("select#inst option:selected").attr(
            'value');
        var custid = $("select#cust option:selected").attr(
            'value');
        var invid = # ("#inv").val()
        if (inst_id.length > 0) {

            $.ajax({
                type: "POST",
                url: "wo_start.php",
                data: {
                    inst_id: inst_id,
                }
                cache: false,
                beforeSend: function() {

                },
                success: function() {

                }
            });


        }
    });
});

I have tried using data: {inst_id:inst_id,custid:custid,invid:invid,} (no update to the table like this)

I also tried data: "inst_id="+inst_id+"&custid="+custid+"&invid="+invid,(this also gives no results.)

Can someone PLEASE look at this jquery and see if I am making a simple error?

Answer 1

Try this format:

data: { inst_id: inst_id, custid: custid, invid: invid },

You can post a JSON object to the server so long as you serialize it and then let the server know the data type.

First you need to define your JSON object:

var postData = { inst_id: inst_id, custid: custid, invid: invid };

Then update your ajax to use the serialized version of that object and let the server know the data type:

$.ajax({
            type: "POST",
            url: "fetch_inst.php",
            data: JSON.stringify(postData),
            contentType: "application/json",
            ..continue the rest of your ajax....