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我有一个文件，form.php，它有一个小表格，用于通过名和姓搜索数据库中的人。该表单使用JavaScript函数通过AJAX将变量发送到search_name.php，并将从mydatabase查询的信息作为表单中的值发送。使用AJAX格式更新数据库
我希望能够使用搜索结果更新#result元素中窗体上的信息。
我试着做一个小例子，没有通过AJAX返回的表单，它的工作，但由于某种原因，我无法做到我的大项目。
任何人都可以请帮忙。我查找了一些示例和信息，但我是AJAX和PHP的新手，无法弄清楚为什么会发生这种情况。
form.php的
<script language="JavaScript" type="text/javascript">
function ajax_post(){
var fn = document.getElementById("first_name").value;
var ln = document.getElementById("last_name").value;
var errorMsg ="";
if (fn==null || fn==""){
errorMsg +="Enter First Name \n";
document.getElementById("first_name").focus();
}
if (ln==null || ln==""){
errorMsg +="Enter Last Name \n";
document.getElementById("last_name").focus();
}
if(errorMsg != ""){
alert(errorMsg);
document.getElementById("first_name").focus();
return false;
}else{
// Create our XMLHttpRequest object
var hr = new XMLHttpRequest();
// Create some variables we need to send to our PHP file
var url = "search_name.php";
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
// Set content type header information for sending url encoded variables in the request
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("result").innerHTML = return_data;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("result").innerHTML = "processing...";
}
}
</script>
</head>
<body>
<div class="left" id="search">
First Name: <input id="first_name" name="first_name" type="text" />
<br /><br />
Last Name: <input id="last_name" name="last_name" type="text" />
<br /><br />
<input name="myBtn" type="submit" value="Search" onClick="javascript:ajax_post();return">
<br /><br />
</div>
<div id="result"></div>
search_name.php
<?php $form_profile = '<form method="POST" action=""><table width="450px"><tr><td><label for="firstname" >First Name: </label></td><td><input type="text" id="first_name" name="first_name" maxlength="50" size="30" value="'.$first_name.'"/></td></tr><tr><td><label for="lastname" >Last Name: </label></td><td><input type="text" id="last_name" name="last_name" maxlength="50" size="30" value="'.$last_name.'"/></td></tr><tr><td><label for="headline">Headline</label></td><td><input type="text" id= "headline" name="headline" maxlength="50" size="30" value="'.$profile_headline.'"/></td></tr></table><input type="submit" id="submit" name="submit" value="Save and Update"></form>'; ?>
<?php
//check if form has been submitted
if(isset($_POST['submit'])){
$first_name= $_POST['first_name'];
$last_name= $_POST['last_name'];
$headline= $_POST['headline'];
$summary= $_POST['summary'];
$title_array= $_POST['title'];
$company_array= $_POST['company'];
$start_month_array= $_POST['start_month'];
$start_year_array= $_POST['start_year'];
$end_month_array= $_POST['end_month'];
$end_year_array= $_POST['end_year'];
if($first_name && $last_name){
//connect to server
$link = mysql_connect("localhost", "root", "########");
if($link){
mysql_select_db("mydatabase",$link);
}
//check if person exists
$exists = mysql_query("SELECT * FROM profile WHERE firstname = '$first_name' AND lastname = '$last_name'") or die ("The query could not be complete.");
if(mysql_num_rows($exists) != 0){
//update
mysql_query("UPDATE profile SET headline='$headline' WHERE firstname = '$first_name' AND lastname = '$last_name'") or die("Update could not be applied");
echo "Success!!";
}else echo "That alumni is not in the database";
}else echo "You must provide a first and last name.";
}
?>
**危险**：您正在使用[an **过时的**数据库API]（http://stackoverflow.com/q/12859942/19068）并应使用[现代替换]（http：// php.net/manual/en/mysqlinfo.api.choosing.php）。你也很容易受到[SQL注入攻击]（http：// bobby-tables。/ **）一个现代的API会让你更容易[防守]（http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php）。 – Quentin