PHP聯系表格不起作用（3）
[英]PHP contact form doesn't work (3)
問題描述
我正在按照本教程http://www.sanwebe.com/2011/12/making-simple-jquery-ajax-contact-form創建php-ajax-jquery聯系表單
看起來它適用於localhost（它顯示發送電子郵件的消息），但它在服務器上不起作用。
你能幫我解決一些問題：
- 在嘗試使用localhost時是否會發送電子郵件？
- 如何修復服務器問題，為什么推送提交按鈕后沒有任何反應？
這是索引文本
<!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Simple Ajax Contact Form</title> <script src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#Submit").click(function() { var proceed = true; //simple validation at client's end //loop through each field and we simply change border color to red for invalid fields $("#contact_form input[required=true], #contact_form textarea[required=true]").each(function(){ $(this).css('border-color',''); if(!$.trim($(this).val())){ //if this field is empty $(this).css('border-color','red'); //change border color to red proceed = false; //set do not proceed flag } }); if(proceed) //everything looks good! proceed... { //get input field values data to be sent to server post_data = { 'user_name' : $('input[name=name]').val(), 'phone_number' : $('input[name=phone2]').val() }; //Ajax post data to server $.post('contact_me.php', post_data, function(response){ if(response.type == 'error'){ //load json data from server and output message output = '<div class="error">'+response.text+'</div>'; }else{ output = '<div class="success">'+response.text+'</div>'; //reset values in all input fields $("#contact_form input[required=true], #contact_form textarea[required=true]").val(''); $("#contact_form #contact_body").slideUp(); //hide form after success } $("#contact_form #contact_results").hide().html(output).slideDown(); }, 'json'); } }); //reset previously set border colors and hide all message on .keyup() $("#contact_form input[required=true], #contact_form textarea[required=true]").keyup(function() { $(this).css('border-color',''); $("#result").slideUp(); }); }); </script> <link href="style.css" rel="stylesheet" type="text/css" /> <!--<link href="/zvonoq/css/bootstrap.min.css" rel="stylesheet" type="text/css" />--> </head> <body> <div class="form-style" id="contact_form"> <div class="form-style-heading">Please Contact Us</div> <div id="contact_results"></div> <div id="contact_body"> <label><span>Name <span class="required">*</span></span> <input type="text" name="name" id="name" required="true" class="input-field"/> </label> <label><span>Phone</span> <input type="text" name="phone2" maxlength="15" required="true" class="tel-number-field long" /> </label> <label> <span> </span><input type="submit" id="Submit" value="Submit" /> </label> </div> </div> </body> </html>
這是contact_me.php
<?php if($_POST) { $to_email = "myemail@gmail.com"; //Recipient email, Replace with own email here //check if its an ajax request, exit if not if(!isset($_SERVER['HTTP_X_REQUESTED_WITH']) AND strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) != 'xmlhttprequest') { $output = json_encode(array( //create JSON data 'type'=>'error', 'text' => 'Sorry Request must be Ajax POST' )); die($output); //exit script outputting json data } //Sanitize input data using PHP filter_var(). $user_name = filter_var($_POST["user_name"], FILTER_SANITIZE_STRING); $user_email = filter_var($_POST["user_email"], FILTER_SANITIZE_EMAIL); $country_code = filter_var($_POST["country_code"], FILTER_SANITIZE_NUMBER_INT); $phone_number = filter_var($_POST["phone_number"], FILTER_SANITIZE_NUMBER_INT); $subject = filter_var($_POST["subject"], FILTER_SANITIZE_STRING); $message = filter_var($_POST["msg"], FILTER_SANITIZE_STRING); //additional php validation if(strlen($user_name)<4){ // If length is less than 4 it will output JSON error. $output = json_encode(array('type'=>'error', 'text' => 'Name is too short or empty!')); die($output); } if(!filter_var($phone_number, FILTER_SANITIZE_NUMBER_FLOAT)){ //check for valid numbers in phone number field $output = json_encode(array('type'=>'error', 'text' => 'Enter only digits in phone number')); die($output); } //email body $message_body = $message."\\r\\n\\r\\n-".$user_name."\\r\\nEmail : ".$user_email."\\r\\nPhone Number : (".$country_code.") ". $phone_number ; //proceed with PHP email. $headers = 'From: '.$user_name.'' . "\\r\\n" . 'Reply-To: '.$user_email.'' . "\\r\\n" . 'X-Mailer: PHP/' . phpversion(); $send_mail = mail($to_email, $subject, $message_body, $headers); if(!$send_mail) { //If mail couldn't be sent output error. Check your PHP email configuration (if it ever happens) $output = json_encode(array('type'=>'error', 'text' => 'Could not send mail! Please check your PHP mail configuration.')); die($output); }else{ $output = json_encode(array('type'=>'message', 'text' => 'Hi '.$user_name .' Thank you for your email')); die($output); } } ?>
1 個解決方案
解決方案1
0 2015-03-06 21:24:22
看起來這樣做比以前更加復雜。 將contact_body更改為<form> （不要忘記更改結束標記）
然后，您可以使用jQuery簡單地執行此操作以發送表單
<?php
$.post("www.someSite.com/someScript.php",
$("#contact_body").serialize())
.error(function(){
//Show alert for not being able to make a connection
})
.success(function(data){
var response = JSON.parse(data); //Get json object from response text
if(response..type != error){
//Show some error
}else{
//All went well
}
});
如果您單擊按鈕並且沒有發生任何事情，那么您的javascript就會出現問題。 打開Web / Javascript控制台以檢查錯誤
PHP聯絡表格無效
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