Php 基于下拉框使用AJAX返回查询结果
我知道这是一个很受欢迎的问题，我已经看过很多例子试图了解AJAX和jQuery 我有一个简单的情况，当change根据下拉框选择发送AJAX请求以获取SQL查询结果时，会有一个下拉框 当从下拉框中选择某个部门时，页面正确加载并调用函数。警报告诉我，但我没有收到任何返回数据。在试图识别问题时，我如何判断getTeachers.php文件是否正在实际运行 网页 用于在服务器上调用getTeacher.php的脚本Php 基于下拉框使用AJAX返回查询结果,php,sql,ajax,Php,Sql,Ajax,我知道这是一个很受欢迎的问题，我已经看过很多例子试图了解AJAX和jQuery 我有一个简单的情况，当change根据下拉框选择发送AJAX请求以获取SQL查询结果时，会有一个下拉框 当从下拉框中选择某个部门时，页面正确加载并调用函数。警报告诉我，但我没有收到任何返回数据。在试图识别问题时，我如何判断getTeachers.php文件是否正在实际运行 网页 用于在服务器上调用getTeacher.php的脚本 <script src="http://localhost/jquery/jqu
<script src="http://localhost/jquery/jquery.min.js">
</script>
<script>
function checkTeacherList(str)
{
var xmlhttp;
if (str=="")
{
document.getElementById("txtTeacher").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtTeacher").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getTeachers.php?q="+str,true);
xmlhttp.send();
alert(str); //To test it is getting this far, which it does
}
</script>
<select name="department_list" id="department_list" onchange="checkTeacherList(this.value);" >
<?php
$options[0] = 'All';
$intloop = 1;
while($row = mysql_fetch_array($department_result))
{
$options[$intloop] = $row['departmentName'];
$intloop = $intloop + 1;
}
foreach($options as $value => $caption)
{
echo "<option value=\"$caption\">$caption</option>";
}
?>
</select>
<div id="txtTeachers">Teacher names</div>
将getTeacher.php页面中的查询更改为。
$sql=从部门名称为“$q”的部门中选择* 我记得我第一次用Jquery请求Ajax时，发现很难找到一个完整的好例子，尤其是一些错误处理的例子。如果后端出现问题，例如数据库不可用，我如何告诉用户 以下是您使用PDO和Jquery重写的代码，包括一些错误处理。我没有使用Mysql扩展，因为它是从最新的PHP版本中删除的。顺便说一句，您的代码可以接受sql注入，删除数据库非常容易：
<!DOCTYPE html>
<html>
<head>
<title>Selectbox Ajax example</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<body>
<div id="error"></div>
<select name="department_list" id="department_list">
<option value="department1">Department 1</option>
<option value="department2">Department 2</option>
</select>
<div id="txtTeachers">Teacher names</div>
<div id="result">
<ul id="list">
</ul>
</div>
<script type="text/javascript">
$(document).ready(function () {
// if user chooses an option from the select box...
$("#department_list").change(function () {
// get selected value from selectbox with id #department_list
var selectedDepartment = $(this).val();
$.ajax({
url: "getTeachers.php",
data: "q=" + selectedDepartment,
dataType: "json",
// if successful
success: function (response, textStatus, jqXHR) {
// no teachers found -> an empty array was returned from the backend
if (response.teacherNames.length == 0) {
$('#result').html("nothing found");
}
else {
// backend returned an array of names
var list = $("#list");
// remove items from previous searches from the result list
$('#list').empty();
// append each teachername to the list and wrap in <li>
$.each(response.teacherNames, function (i, val) {
list.append($("<li>" + val + "</li>"));
});
}
}});
});
// if anywhere in our application happens an ajax error,this function will catch it
// and show an error message to the user
$(document).ajaxError(function (e, xhr, settings, exception) {
$("#error").html("<div class='alert alert-warning'> Uups, an error occurred.</div>");
});
});
</script>
</body>
</html>
<?php
// we want that our php scripts sends an http status code of 500 if an exception happened
// the frontend will then call the ajaxError function defined and display an error to the user
function handleException($ex)
{
header('HTTP/1.1 500 Internal Server Error');
echo 'Internal error';
}
set_exception_handler('handleException');
// we are using PDO - easier to use as mysqli and much better than the mysql extension (which will be removed in the next versions of PHP)
try {
$password = null;
$db = new PDO('mysql:host=localhost;dbname=iobserve', "root", $password);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// note the quote thing - this prevents your script from sql injection
$data = $db->query("SELECT teacherName FROM departments where departmentName = " . $db->quote($_GET["q"]));
$teacherNames = array();
foreach ($data as $row) {
$teacherNames[] = $row["teacherName"];
}
// note that we are wrapping everything with json_encode
print json_encode(array(
"teacherNames" => $teacherNames,
"anotherReturnValue" => "just a demo how to return more stuff")
);
} catch (PDOException $e) {
echo 'ERROR: ' . $e->getMessage();
}
您需要以单个字符串发送响应。因此，创建一个查询返回的所有教师的字符串
然后在ajax部分，拆分这个字符串 为了测试文件是否运行，您可以使用一些硬编码的虚拟数据来响应，而不是运行可能失败的代码。谢谢-结果证明，我需要将完整路径放入php文件。我在WordPress上运行它，它一定很难找到它。如果您不希望每个人都可以删除您的数据库，请将mysql\u real\u escape\u字符串添加到您的select查询中。谢谢-问题是缺少分号。。。。。部门名称='.$q.'；；首先，我非常感谢你花了这么多时间回答这个问题。我同意刚开始时很难找到好的例子。事实上，我已经让它工作起来了，并且似乎对它有了更好的理解，但我会仔细检查您发送的内容以进一步改进它。我很想知道什么是SQL注入，但我也会查一下。我知道我需要离开MySQL。。。要学的东西太多了。再次感谢。
<?php
// we want that our php scripts sends an http status code of 500 if an exception happened
// the frontend will then call the ajaxError function defined and display an error to the user
function handleException($ex)
{
header('HTTP/1.1 500 Internal Server Error');
echo 'Internal error';
}
set_exception_handler('handleException');
// we are using PDO - easier to use as mysqli and much better than the mysql extension (which will be removed in the next versions of PHP)
try {
$password = null;
$db = new PDO('mysql:host=localhost;dbname=iobserve', "root", $password);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// note the quote thing - this prevents your script from sql injection
$data = $db->query("SELECT teacherName FROM departments where departmentName = " . $db->quote($_GET["q"]));
$teacherNames = array();
foreach ($data as $row) {
$teacherNames[] = $row["teacherName"];
}
// note that we are wrapping everything with json_encode
print json_encode(array(
"teacherNames" => $teacherNames,
"anotherReturnValue" => "just a demo how to return more stuff")
);
} catch (PDOException $e) {
echo 'ERROR: ' . $e->getMessage();
}