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我ajax_form.php頁：顯示在同一頁面Ajax內容沒有頁面加載
<html><head>
<meta content="text/html;charset=utf-8" http-equiv="Content-Type">
<meta content="utf-8" http-equiv="encoding">
<script>
function showUser(form, e) {
e.preventDefault();
var xmlhttp;
var submit = form.getElementsByClassName('submit')[0];
var sent = document.getElementsByName('sent')[0].value || '';
var id = document.getElementsByName('id')[0].value || '';
if (sent==""){
document.getElementById("txtHint").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(e) {
if (xmlhttp.readyState==4 && xmlhttp.status==200){
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open(form.method, form.action, true);
xmlhttp.send('sent='+sent+'&id='+id+'&'+submit.name+'='+submit.value);
}
</script>
<form action="ajax_test.php" method="POST">
Enter the sentence: <input type="text" name="sent"><br>
<input type="submit" class="submit" name="insert" value="submit" onsubmit="showUser(this, event)">
</form>
<br>UPDATE <br>
<form action="ajax_test.php" method="POST" onsubmit="showUser(this, event)">
<pre>
Enter the ID : <input type="text" name="id"><br>
Enter the sentence: <input type="text" name="sent"><br>
</pre>
<input type="submit" class="submit" value="submit" name="update" >
</form> <br>
<div id="txtHint">
<b>Person info will be listed here.</b>
</div>
</body>
</html>
和ajax_test.php是：
<html><head>
<meta content="text/html;charset=utf-8" http-equiv="Content-Type">
<meta content="utf-8" http-equiv="encoding">
</head> <body >
<?php
// $q = $_POST["q"];
// you never process the $q var so i commented it
if (isset($_POST['insert']) && $_POST['insert'] !== '') {
echo "Operation: Insert","<br>";
$s = $_POST['sent'];
$flag = 0;
echo "Entered sentence : $s";
if (preg_match_all('/[^=]*=([^;@]*)/',
shell_exec("/home/technoworld/Videos/LinSocket/client '$s'"),
$matches)){ //Values stored in ma.
$x = (int) $matches[1][0]; //optionally cast to int
$y = (int) $matches[1][1];
}
echo "<br>",
"Positive count :$x",
"<br>",
"Negative count :$y",
"<br>";
//---------------DB stuff --------------------
$con = mysqli_connect('127.0.0.1:3306', 'root', 'root', 'test');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql1 = "INSERT INTO table2
(id,sent,pcount,ncount,flag)
VALUES
('','".$_POST['sent']."',' $x ','$y','$flag')";
if (mysqli_query($con, $sql1)) {
echo "1 record added";
} else {
die('Error: ' . mysqli_error($con));
}
mysqli_close($con);
}
// -------------------------------UPDATE --------------------------
if (isset($_POST['update']) && $_POST['update'] !== '') {
echo "Operation: update", "<br>";
// you say update but you are actually inserting below
$s = $_POST['sent'];
$flag = 1;
echo "Entered sentence : $s";
if (preg_match_all('/[^=]*=([^;@]*)/',
shell_exec("/home/technoworld/Videos/LinSocket/client '$s'"),
$matches)) //Values stored in ma.
{
$x = (int) $matches[1][0]; //optionally cast to int
$y = (int) $matches[1][1];
}
echo "<br>",
"Positive count :$x",
"<br>",
"Negative count :$y",
"<br>";
//---------------DB stuff --------------------
$con = mysqli_connect('127.0.0.1:3306', 'root', 'root', 'test');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql1 = "INSERT INTO table2
(id,sent,pcount,ncount,flag)
VALUES
('','".$_POST['sent']."',' $x ','$y','$flag')"; // error here again $_POST[id] should be $_POST['id'] with quotes
if (mysqli_query($con, $sql1)) {
echo "1 record added";
} else {
die('Error: ' . mysqli_error($con));
}
mysqli_close($con);
}
?></html > </body >
在Form1我已經把函數調用按鈕單擊事件，它工作正常。但在按鈕上單擊它加載頁面並重定向到ajax_test.php。我們可以說它正確使用ajax嗎？
在第二種形式中，我保留了函數調用的形式本身，並按照腳本中的要求進行了編碼。但按鈕上點擊行動發生。函數調用或其他錯誤是錯誤的嗎？
如何在兩種情況下顯示沒有頁面加載（刷新）的結果？
@StevenMoseley：對不起，我簽了一段時間！我來檢查一下！ – user123
你不會偶爾使用IE嗎？在那裏你可能需要一個e.returnValue = false而不是e.preventDefault（）。另一個使用像jQuery這樣的庫的原因隱藏了很多瀏覽器的不兼容性。它也會簡化你的代碼。 – mvw
@mvw：謝謝，我希望'e.preventDefault（）; e.returnValue = false;即使是IE也不會造成問題！ – user123