输入其中一个字段时填充表单输入元素（文本类型），而不刷新HTML表单。 [英] Populate form input elements(text-typed) when one of the field is entered, without refreshing the HTML form.

问题描述

输入customer_phone_number字段后，我无法在html表单中填写客户名称，电子邮件和地址字段。

当我输入客户电话号码并将光标移出时我希望我的脚本能够将电话号码发送到php文件并从数据库中的customers表中获取数据并将其发送到脚本和脚本以填充html输入字段而不刷新页面，以便我可以填写其他字段并提交表单。



我试过的代码没有提取任何数据。请帮助我让我的代码正常工作。



我尝试过：

I am unable to populate customer name, email and address fields in the html form after entering the customer_phone_number field.
when i enter the customer phone number and move the cursor out i want my script to send phone number to php file and fetch data from customers table in the database and send it to the script and the script to populate the html input fields without refreshing the page so that i can fill other fields and submit the form.

the code that i tried is not fetching any data. please help me getting my code working.

What I have tried:

this is my ajax code

function showUser(str) {
  if (str=="") {
    document.getElementById("full_name").innerHTML="";
	document.getElementById("email").innerHTML="";
	document.getElementById("address").innerHTML="";
	
    return;
  }
  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp=new XMLHttpRequest();
  } else { // code for IE6, IE5
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function() {
    if (this.readyState==4 && this.status==200) {
		var myObj = JSON.parse(this.responseText);
        document.getElementById("full_name").innerHTML = myObj.full_name;
        document.getElementById("email").innerHTML=myObj.email;
		document.getElementById("address").innerHTML=myObj.address;
    }
  }
  xmlhttp.open("GET","getinfo.php?q="+str,true);
  xmlhttp.send();
}

this is my html code

		<form name="testform">
		
		<h1> customer information </h1>
		Order_no:
		<input type="text" name = "order_no" id="order_no"><br>
		<!-- i was able to auto generate the order_no field with the last record table in the database -->
		customer_phone_number:
		<input type="text" name="customer_phone_number" id="customer_phone_number" onpointermove="showUser(this.value)"><br>
		customer_full_name:
		<input type="text" name="full_name" id="full_name"><br>
		customer_email:
		<input type="text" name="email" id="email"><br>
		customer_address:
		<input type="textarea" name="address" id="address"><br>
		
	   <!-- i will be taking many other information for my application-->
		<input type="submit" name="save" value = "save">

</form>

this is my php code getinfo.php

<?php

	$host = 'localhost';
	$user = 'root';
	$pass = '';
	$dbname = 'yukba';
	
	$conn = mysqli_connect( $host,$user,$pass,$dbname );
	$q = $_GET['q'];
	$myObj->full_name = "";
	$myObj->email = "";
	$myObj->address = "";
	
	if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
	}
	
	$sql = "SELECT FULL_NAME,EMAIL,ADDRESS FROM customers WHERE CUSTOMER_PHONE_NUMBER= {$q}";
	$result = mysqli_query($conn, $sql);
	
	if (mysqli_num_rows($result) > 0) {
    
    while($row = mysqli_fetch_assoc($result)) {
	
	$myObj->full_name = $row['FULL_NAME'];;
	$myObj->email = $row['EMAIL'];
	$myObj->address = $row['ADDRESS'];
	
	$myJSON = json_encode($myObj);

	echo $myJSON;
        
    }
	} else {
    echo "0 results";
	}

?>

推荐答案

host = ' localhost';

host = 'localhost';

user = ' 根;

user = 'root';

pass = ' ';

pass = '';

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