供参考，我使用的是mySQL数据库。这必须在没有页面刷新的情况下完成（因此需要使用AJAX完成[我正在使用jQuery来实现我的AJAX功能]）。例如，有一个人有一个名字，姓氏，他们被分配到最少一个工作或最多2个工作。有2个表：
第一个表从作业表中获取id并将其用作外键。
表“tbl_person”：
+------+-------------+------------+--------+--------+
| p_ID | p_firstName | p_lastName | p_job1 | p_job2 |
+------+-------------+------------+--------+--------+
| 1 | Mike | Jefferson | 3 | 1 |
| 2 | Jack | Frost | 2 | |
+------+-------------+------------+--------+--------+
表“tbl_job”
+--------+------------+
| job_ID | job_Name |
+--------+------------+
| 1 | Clerk |
| 2 | Programmer |
| 3 | Mechanic |
+--------+------------+
我想用tbl_job表中的job_Name显示tbl_person表。到目前为止，PHP代码本身正确地显示了页面刷新。但是，使用我当前的jQuery代码，只显示来自tbl_person的数据。
jQuery当前显示的示例（例如，使用一个显示所有条目）：
+------+-----------+---+---+----+------+
| Mike | Jefferson | 3 | 1 |Edit|Delete|
+------+-----------+---+---+----+------+
php代码的功能示例/我想要jQuery代码执行的操作（例如，使用一个显示所有条目）：
+------+-----------+----------+-------+----+------+
| Mike | Jefferson | Mechanic | Clerk |Edit|Delete|
+------+-----------+----------+-------+----+------+
代码的主要部分如下。我想在html表中显示我正在登录控制台的内容，以便它显示在网页上（见上表）：
$.ajax({
url: "person.php",
type: 'POST',
data: data,
success: function (data) {
console.log($("#job1 option:selected").text());
console.log($("#job2 option:selected").text());
},
error: function () {
//insert error alert
}
});
函数pdo文件未包含（包含db函数）
带有输入数据和显示格式的PHP文件（html结构）： person.php
//below code posts new info to database
<?php
$pdo = new functions();
$getPerson = $pdo->getPerson("");
$PersonList = $pdo->getPersonList();
if(isset($_POST["p_ID"])){
if($_POST["p_ID"] != ''){
$pdo->updatePerson($_POST);
}else{
$pdo->addPerson($_POST);
}
exit();
}
if(isset($_GET["id"])) {
$pdo->deletePerson($_GET["id"]);
}
?>
//html form for entering info (this should pop up with jQuery code when clicking new entry button
<div id="personform" title="Person Form">
<form name="personedit" id="personedit" method="post" action="person.php">
<!--//hidden section for the id to pass for the record update-->
<input name="p_ID" type="hidden" value="" />
<fieldset>
<legend>People</legend>
<table>
<tr>
<th><label for="p_firstName">First Name:</label><br /></th>
<td><input type="text" class="name" name="p_firstName" value="" size="18" maxlength="25"/></td>
<th><label for="p_lastName">Last Name:</label><br /></th>
<td><input type="text" class="name" name="p_lastName" value="" size="18" maxlength="25"/></td>
</tr>
<tr>
<th>First Job:</th>
<td>
<select name="p_job1" id="job1" class="name">
<option value=""></option>
<?php
$getJobs = $pdo->getJobs();
foreach($getJobs as $row){
if($row["job_ID"] == $getPerson["p_job1"]){
$selected = "selected = selected";
}else{
$selected = "";
}
echo "<option value='" . $row['job_ID'] . "' $selected>" . $row['job_name'] . "</option>";
}
?>
</select>
</td>
<th>Second Job:</th>
<td>
<select name="p_job2" id="job2" class="name">
<option value=""></option>
<?php
foreach($getJobs as $row){
if($row["job_ID"] == $getPerson["p_job2"]){
$selected = "selected = selected";
}else{
$selected = "";
}
echo "<option value='" . $row['job_ID'] . "' $selected>" . $row['job_name'] . "</option>";
}
?>
</select>
</td>
</tr>
</table>
</fieldset>
</form>
</div>
/*html code to display the person table data in table format with the job name
displayed (not the id) */
<a class="add" href="person.php">New Entry</a>
<h1>Person List</h1>
<div class="float-left">
<table id="tbllist">
<tr>
<th>Last Name</th>
<th>First Name</th>
<th>First Job</th>
<th>Second Job</th>
<th colspan="2">Action</th>
</tr>
<?php
foreach ($PersonList as $row) {
echo "<tr class='fadeaway" . $row['p_ID'] . "'>";
echo "<td><input type='hidden' id='p_lastName".$row['p_ID']."' value='".$row['p_lastName']."'>".$row['p_lastName']."</td>";
echo "<td><input type='hidden' id='p_firstName".$row['p_ID']."' value='".$row['p_firstName']."'>".$row['p_firstName']."</td>";
echo "<td><input type='hidden' id='job_Name1".$row['p_ID']."' value='".$row['p_job1']."'>".$row['job_Name1']."</td>";
echo "<td><input type='hidden' id='job_Name2".$row['p_ID']."' value='".$row['p_job2']."'>".$row['job_Name2']."</td>";
echo "<td><a href='#' class='edit' id='".$row['p_ID']."'>Edit</a></td>";
echo "<td><a href='#' class='delete' id='".$row['p_ID']."'>Delete</a></td>";
echo "</tr>";
}
?>
</table>
</div>
Javascript代码的一部分
//this section is in a separate js file
// JavaScript Document
function() {
var data = $($(this).data('formID')).closest("form").serialize();
if($($(this).data('formID')).valid()){
var ID = $(this).data('ID'); //gets ID of table row for editing
//This function finds the class name of each input for use in the add section
var inputname = $(".name").map(function() {
r = $(this).attr("name");
return r;
}).get();
//This function finds the value of each input for use in the add section
var inputval = $(".name").map(function() {
p = $(this).val();
return p;
}).get();
//code to change array positioning so that the table is printed correctly.
/*AJAX code for name value (this is what needs to be changed) */
$.ajax({
url: "person.php",
type: 'POST',
data: data,
success: function (data) {
},
error: function () {
alert('Sorry, there was a problem!');
}
});
$.ajax({
'url': '',
'type': 'POST',
'dataType': 'text',
'data': data,
'success': function(data){
if (action == "edit") {
var tblval = "";
//loops through the number of input names and adds a table value for each one
for (i = 0; i < inputname.length; i++) {
console.log(inputval[i]);
var val = inputval[i];
var tblval = tblval + "<td><input type='hidden' id="+inputname[i]+ID+" value='"+val+"'>"+val+"</td>";
}
edit = $("#tbllist .fadeaway"+ID+"").html(tblval+"<td><a href='#' class='edit' id='"+ID+"'>Edit</a></td>"+"<td><a href='#' class='delete' id='"+ID+"'>Delete</a></td>");
}
}
});
}
}
我对AJAX和jQuery相当新，所以这些代码可能不是最佳实践。
答案 0 :(得分：0)
我编写的代码还有其他问题，但我通过更改此部分解决了这个问题：
var inputval = $(".name").map(function() {
p = $(this).val();
return p;
}).get();
到此：
var inputval = $(".name").map(function() {
if (!$(this).is("select")) {
p = $(this).val();
} else {
p = $(this).find("option:selected").text();
}
return p;
}).get();