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只發送的第一個值
我做的AJAX功能使用表單和輸入類型=隱藏插入做到數據庫,但是當我在同一頁面放置超過形式,它始終是在第一種形式不止一種形式以相同的輸入ID與阿賈克斯
這裏僅舉第一個值是代碼:
Ajax的功能
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById("ajaxDiv");
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
var in1 = document.getElementById("in1").value;
var queryString1 = "?in1=" + in1;
//ajaxRequest.open("GET", "/sites/all/php/check.php" + queryString, true);
ajaxRequest.open("GET", "/sites/all/php/check.php" + queryString1, true);
ajaxRequest.send(null);
}
//–>
</script>
這裏是HTML表單
<form name="myForm1">
<input type="hidden" value="1" id="in1">
<a onclick="ajaxFunction()" class="folloo"></a>
<form name="myForm2">
<input type="hidden" value="2" id="in1">
<a onclick="ajaxFunction()" class="folloo"></a>
<form name="myForm3">
<input type="hidden" value="3" id="in1">
<a onclick="ajaxFunction()" class="folloo"></a>
</form>
</form>
</form>
這裏是check.php代碼做DB查詢
<?php
//Connect to MySQL Server
//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","admin","123456") or die('Cannot connect to the database
because: ' . mysql_error());
//specify database ** EDIT REQUIRED HERE **
mysql_select_db("dbname") or die("Unable to select database");
//select which database we're using
// Retrieve data from Query String
$in1 = $_GET['in1'];
// Escape User Input to help prevent SQL Injection
$in1 = mysql_real_escape_string($in1);
//Build and run a SQL Query on our MySQL tutorial
if($in1){
mysql_query("INSERT INTO test (name)
VALUES ('" . $in1. "')");
}
?>
看跌當我點擊任何鏈接,它總是發送第一個值。
CAH誰能告訴mewhat是這裏的問題?
document.getElementsByName(「」)。value返回未定義的值 –
@ManMann:'getElementsByName'返回**元素**,複數形式爲'NodeList'。 'value'是*個人*元素的屬性。 (另外,我假設你沒有像上面顯示的那樣向它傳遞一個空字符串?)如果你給它一個帶有元素名稱的字符串,然後查看結果列表的「長度」,並索引到它與'[0]','[1]'等,你應該能夠找到每個這些'價值'屬性。 –
:我做了更改,但仍返回undefined值:( –