使用AJAX更改PHP db_query參數
[英]Changing a PHP db_query parameter with AJAX
問題描述
假設我有這段代碼：
$templatesf = $DB->query('SELECT * FROM templates WHERE category="something"');
有沒有辦法使用javascript / AJAX更改“內容”？ 例如：
function changesomething(selse){
if (selse == '1'){
something = 'this'
}else{
something = 'that'
}
}
2 個解決方案
解決方案1
0 2014-04-15 17:21:22
您需要放置一個變量而不是該變量，然后將ajax變量傳遞給該方法。
您將使用$ _POST捕獲變量並將其放置，而不是放置先前的變量。
您也可以使用默認值（以防萬一，或者傳遞的POST錯誤）。
解決方案2
0 已采納 2014-04-15 17:49:38
以下將是您的輸入頁面，從這里您將調用ajax處理程序來處理something 。
<html>
<head>
<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 and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var somedata = document.getElementById('something').value;
var queryString = "?something=" + somedata ;
ajaxRequest.open("GET", "ajax-example.php" +
queryString, true);
ajaxRequest.send(null);
}
</script>
</head>
<body>
<form name='myForm'>
Something: <input type='text' id='something' /> <br />
<input type='button' onclick='ajaxFunction()'
value='Query MySQL'/>
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>
下面是PHP代碼來處理您的請求並處理您的something
<?php
$dbhost = "localhost";
$dbuser = "dbusername";
$dbpass = "dbpassword";
$dbname = "dbname";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$something = $_GET['something'];
// Escape User Input to help prevent SQL Injection
$something = mysql_real_escape_string($something);
//build query
$query = "SELECT * FROM ajax_example WHERE something = '$something'";
$qry_result = mysql_query($query) or die(mysql_error());
//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>something</th>";
$display_string .= "<th>other</th>";
$display_string .= "<th>blahblah</th>";
$display_string .= "</tr>";
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string .= "<tr>";
$display_string .= "<td>$row[name]</td>";
$display_string .= "<td>$row[something]</td>";
$display_string .= "<td>$row[other]</td>";
$display_string .= "<td>$row[blahblah]</td>";
$display_string .= "</tr>";
}
echo "Query: " . $query . "<br />";
$display_string .= "</table>";
echo $display_string;
?>
使用AJAX更改PHP函數中的參數並重新加載
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