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javascript - 使用AJAX形式更新数据库

转载 作者:行者123 更新时间:2023-11-29 13:17:23 25 4
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我有一个文件 form.php,其中有一个小表单,用于按名字和姓氏在数据库中搜索人员。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。

我希望能够使用搜索结果更新 #result 元素中表单上的信息。

我尝试做一个没有通过 AJAX 返回表单的小示例,它有效,但由于某种原因我无法在我的更大项目中执行此操作。

任何人都可以帮忙吗?我已经查找了示例和信息,但我对 AJAX 和 PHP 很陌生,无法弄清楚为什么会发生这种情况。

form.php

<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.";


}

?>

最佳答案

正如 Timmy 提到的,没有发布任何提交值(只有通过表单触发发布时才会自动发生)。另外,当您发送“firstname”时,您会尝试获取 $_POST['first_name'](last_name 与 lastname 也是如此)。

当您使用 JavaScript/AJAX 时,使用某种开发人员工具非常重要。我个人使用 Chrome 开发者工具(在 Chrome 中按 F12 https://developers.google.com/chrome-developer-tools/ )。这将向您显示请求/响应的实际情况,以便您可以找出问题所在。根据您的前端所做的工作,我快速重写了您要发布到的 PHP 脚本:

<?php
//check if form has been submitted
if(isset($_POST['firstname']) || isset($_POST['lastname'])){

$first_name= $_POST['firstname'];
$last_name= $_POST['lastname'];

/*
$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'];
*/


//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 LIKE $first_name.'%' AND lastname LIKE $last_name.'%'") or die ("The query could not be completed.");
if(mysql_num_rows($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.";
}

?>

我修复了错误的变量名称,并注释掉了目前未发送的变量名称。我还更新了您的 MySQL 查询以使用 LIKE 字符串比较函数,这对于搜索来说要好得多。这样,如果有人不知道姓氏或只知道姓氏的一部分,他们仍然可以完成查找。有关字符串比较函数的更多信息,请参见:http://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html 。复制并粘贴代码现在应该可以解决您的问题。

关于javascript - 使用AJAX形式更新数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21296563/

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