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这是我的java脚本文件：

$('#addSchoolForm').trigger("reset");

//$(document).ready(function()  {
$(function() {
    $("#dialog").dialog({
        autoOpen: false,
        maxWidth:600,
        maxHeight: 350,
        width: 500,
        height: 300,
    });

    $("#addSchool").on("click", function() {
        $("#dialog").dialog("open");
    });

    $("#addSchoolForm").submit(function(e) { 
        e.preventDefault();
        $("#dialog").dialog("close")
        var postData = jQuery(this).serialize();
        $.ajax({
            type: "POST",
            url: "AddSchools.php",
            data: postData,
            success: function(data){
                alert(data); }
        });   
    });

    $("#editSchool").submit(function(e) { 
        e.preventDefault();
        var editData = jQuery(this).serialize();
        $.ajax({
            type: "POST",
            url: "GetSchoolID.php",
            data: editData,
            dataType: 'json',
            success: function(data){


                var schoolID = $.parseJSON(data);
                alert("success");
                alert(schoolID.name);

                //alert(data["json"]);
                //alert(data); 

                //document.addSchoolForm[sname].value = data[0].name;
                //document.addSchoolForm[abbrev].value = data[abbrev];
                //document.addSchoolForm[abbrev].value = data[0].abbrev;
            }

            alert(schoolID.name); 
        });
        //$("#dialog").dialog("open");
    });
})

这是我的 get schoolID php 文件

<?php
$school_id = $_POST['school_id'];


$db = mysqli_connect("localhost", "root", "imagroup123","mytrack");

if(!$db){
    exit("Error in database connection");
    echo("couldn't connect to database");
}
else{
    $q = "SELECT * FROM `School` WHERE `SchoolID`='$school_id'";
    $schoolresults = mysqli_query($db,$q);

    $row = mysqli_fetch_assoc($schoolresults);
    $school["name"] = $row['SchoolLong'];
    $school["abbrev"] = $row['SchoolShort'];

    echo json_encode($school);

    }


?>

当我刚刚使用 jsonlint.com 测试 php 文件时，我得到了一个正确的 json 对象，但它没有通过 javascript 文件进行。我对此很陌生，所以我很讨厌这个问题。我还想将数据添加到表单值，然后打开对话框表单。

4

2 回答 2

1

改变：

        success: function(data){
            var schoolID = $.parseJSON(data);

至：

        success: function(schoolID){

因为在您指定时$.ajax自动调用.$.parseJSON()dataType: 'json'

于 2013-11-09T00:33:42.153 回答
0

这是我一直使用的 php 后端结构：

$query = " SELECT * 
           FROM school 
           WHERE SchoolID = $school_id;
$result = mysqli_query($cxn, $query) or die ("could not query database 1");

if (mysqli_num_rows($result) > 0)
{
    $row = mysqli_fetch_array($result);
    $variablestopass = array
    (
           'schoolname' => $row['SchoolLong'],
         'schoolabbrev' => $row['SchoolShort'],
     );
  echo json_encode($variablestopass);}
else
  { echo "Error selection id"; }

这里有一些 js 来调用它并阅读它：

$.ajax({
        type: 'POST',
         url: 'thenameofyourfile.php';
        data: {schoolid: schoolid},
        dataType: 'json'
        })
         .done( function() { alert( "Got it!"" );
                             Do other stuff here
                            })
         .fail(function(jqXHR, textStatus, errorThrown){
          console.log(jqXHR.responseText, textStatus, errorThrown);
    });
于 2013-11-09T00:34:04.027 回答