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How to implement ajax form submission in php

藏色散人
藏色散人Original
2021-09-19 10:23:392652browse

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.

How to implement ajax form submission in php

#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 = &#39;name=&#39;+ name + &#39;&email=&#39; + email + &#39;&phone=&#39; + 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
                    $(&#39;#contact_form&#39;).html("<div id=&#39;message&#39;></div>");
                    $("#message").html("<p>联系方式已成功提交!</p>")
                            .append("<p>Script design</p>")
                            .hide()
                            .fadeIn(1500, function() {
                                $(&#39;#message&#39;).append("<img id=&#39;checkmark&#39; src=&#39;yes.ico&#39; />");
                            });
                }
            })
            return false;
        })
    })
</script>

php

print_r($_POST);
echo $_POST[&#39;name&#39;].&#39;<br>&#39;."i&#39;m ok!~";die;

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