How to implement ajax form submission in php
How to implement ajax to submit form form in php: 1. Create an HTML and PHP file and create the form form; 2. Pass "$.ajax({type: "post",url:"text7.php ",data:dataString...})" code can be used to submit the form.
#The operating environment of this article: Windows7 system, PHP7.1, Dell G3 computer.
How to implement ajax form submission in php?
Submit the form through php jq ajax
The code is as follows:
html
<div id="contact_form"> <form name="contact" method="post" > <label for="name" id="name_label">姓名</label> <input type="text" name="name" id="name" size="30" value="" class="text-input" /> <label class="error" for="name" id="name_error">此项必填</label> <label for="email" id="email_label">您的Email</label> <input type="text" name="email" id="email" size="30" value="" class="text-input" /> <label class="error" for="email" id="email_error">此项必填</label> <label for="phone" id="phone_label">您的联系电话</label> <input type="text" name="phone" id="phone" size="30" value="" class="text-input" /> <label class="error" for="phone" id="phone_error">此项必填</label> <br /> <input type="button" name="submit" class="button" id="submit_btn" value="我要发送" /> </form> </div>
js
<script language="javascript"> $(function(){ $(".error").hide(); $(".button").click(function(){ var name= $("#name").val(); if(name==""){ $("#name_error").show(3600); $("#name_error").focus(); return false; //return false 的作用是为了让input框效果是一次只出现一个 //当type为submit的时候,不return false 的话就会提交form表单而没有效果 //当type为button的时候,不用return false也可以显示效果 //原话:而只有当出现错误,即为空时,错误才会出现,因为有 return false 的作用,每次仅会出现一个错误。 } var email= $("#email").val(); if(email==""){ $("#email_error").show(3600); $("#email_error").focus(); return false; } var phone= $("#phone").val(); if(phone==""){ $("#phone_error").show(3600); $("#email_error").focus(); return false; } var dataString = 'name='+ name + '&email=' + email + '&phone=' + phone; //document.write(dataString); $.ajax({ type: "post", url:"text7.php", data:dataString, success:function(mag){ // alert(mag);return; //ajax提交form表单php里无法看post值,只能在回调函数里面打印,下面为成功之后的返回效果 $('#contact_form').html("<div id='message'></div>"); $("#message").html("<p>联系方式已成功提交!</p>") .append("<p>Script design</p>") .hide() .fadeIn(1500, function() { $('#message').append("<img id='checkmark' src='yes.ico' />"); }); } }) return false; }) }) </script>
php
print_r($_POST); echo $_POST['name'].'<br>'."i'm ok!~";die;
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