我想从php数组中获取数据并在同一页面上显示。如何 使用搜索框从php数组导入数据。此代码无法正常工作。 这段代码的错误是什么?
foodstore.js
var xmlHttp = createXmlHttpRequestObject();
function createXmlHttpRequestObject(){
var xmlHttp;
if(window.ActiveXObject){
try{
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}catch(e){
xmlHttp = false;
}
} else{
try{
xmlHttp = new XMLHttpRequest();
}catch(e){
xmlHttp = false;
}
}
if(!xmlHttp)
alert("cant create that object hoss!");
else
return xmlHttp;
}
function process(){
if(xmlHttp.readyState==0|| xmlHttp.readyState==4){
food = encodeURIComponent(document.getElementById("userInput").value);
xmlHttp.open("GET","foodstore.php?food="+food,true);
xmlHttp.onreadystatechange = handleServerResponse;
xmlHttp.send(null);
}else{
setTimeout('process()',1000);
}
}
function handleServerResponse(){
if(xmlHttp.readyState==4){
if(xmlHttp.status==200){
xmlResponse = xmlHttp.responseXML;
xmlDocumentElement = xmlResponse.documentElement;
message = xmlDocumentElement.firstChild.data;
document.getElementById("underInput").innerHTML ='<span style="color:blue">'+message+'</span>';
setTimeout('process',1000);
}else{
alert('Something went wrong!');
}
}
}
foodstore.php
<?php
header('Content-Type: text/xml');
echo '<?xml version="1.0" encoding="UTF-8" standalone="yes"?>';
echo '<response>';
$food = $_GET['food'];
$foodArray = array('tuna','bacon','beef','loaf','ham');
if(in_array($food,$foodArray))
echo 'We do have '.$food'!';
elseif($food =='')
echo 'Enter a food you want to buy';
else
echo 'Sorry we don't sell it '.$food'!';
echo '</response>';
?>
的index.html
<html><head>
<script type="text/javascript" src="foodstore.js"></script>
</head>
<body onload="process()">
<h3>The foods </h3>
Order your foods:
<input type="text" id="Userinput"></input>
<div id="underInput"></div>
</body>
</html>
如何通过搜索框搜索来显示数组数据
答案 0 :(得分:1)
我使用jquery更改了代码,简单。你可以尝试一下。
<强>的index.html 强>
<html>
<head>
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script>
$(function()
{
$("#Userinput").keyup(function()
{
process();
});
$("#Userinput").keydown(function()
{
process();
});
$("#Userinput").focus(function()
{
process();
});
$("#Userinput").change(function()
{
process();
});
});
function process() {
var input_food = $("#Userinput").val();
$.ajax({
type: "GET",
url: "foodstore.php",
data: {food: input_food},
success: function(message)
{
$("#underInput").html('<span style="color:blue">' + message + '</span>');
},
error: function()
{
$("#underInput").html('<span style="color:red">Some error occured</span>');
}
});
}
</script>
</head>
<body >
<h3>The foods </h3>
Order your foods:
<input type="text" id="Userinput" ></input>
<div id="underInput"></div>
</body>
</html>
<强> foodstore.php 强>
<?php
if (!empty($_GET['food']))
{
$food = $_GET['food'];
$foodArray = array('tuna', 'bacon', 'beef', 'loaf', 'ham');
if (in_array($food, $foodArray))
echo "We do have " . $food . "!";
elseif ($food == '')
echo "Enter a food you want to buy";
else
echo "Sorry we don't sell it " . $food . "!";
}
else
{
echo "Enter a food you want to buy";
}
?>
我认为它很简单,如果你知道jquery。并且在php中有一个简单的错误你没有逃避(不要)中的额外单引号所以我使用双引号来表示echo语句。复制粘贴并告诉你这是不是你想要的。不管怎样怀疑。