我有一个自动完成的jQuery菜单,它从MySQL数据库输出我拥有的所有用户的名字。我正在尝试将每个选项链接到正确的配置文件。为此,URL类似于:/ profile.php?id = 341,341代表所选用户的ID。
唯一的问题是,当我尝试输入给定用户的ID时,所有用户的ID都显示在URL中......并且我只想要所选用户的ID!
我已尝试使用PHP,但我不知道要添加到以下行中的内容以使其正常工作。
$req = mysql_query("select id, Username, EmailAddress from ***");
应该是WHERE Username ='username'....最后,我知道我应该尝试别的东西,没有PHP,但我只是想以这种方式测试它!谢谢!
<input type="text" name="course" id="course" />
<script type="text/javascript" src="jquery.js"></script>
<script type='text/javascript' src='jquery.autocomplete.js'></script>
<link rel="stylesheet" type="text/css" href="jquery.autocomplete.css" />
<script type="text/javascript">
$().ready(function() {
$("#course").autocomplete("/test/test2.php", {
selectFirst: false,
formatItem: function(data, i, n, value) {
//make the suggestion look nice
return "<font color='#3399CC'>" + value.split("::")[0] + "</font>";
},
formatResult: function(data,value) {
//only show the suggestions and not the URLs in the list
return value.split("::")[0];
}
}).result(function(event, data, formatted) {
//redirect to the URL in the string
var pieces = formatted.split("::");
window.location.href = '/profile.php?id='+
<?php
mysql_connect ("***", "***","***") or die (mysql_error());
mysql_select_db ("***");
$req = mysql_query("select id, Username, EmailAddress from ***");
while($dnn = mysql_fetch_array($req))
{
echo $dnn['id'];
}
?>
;
console.log(data);
console.log(formatted);
});
});
</script>
答案 0 :(得分:1)
您的MySQL查询对数据库中的每个用户都是如此,因此它返回所有用户。如果你想转到“foo”的个人资料,你需要告诉数据库只提取“foo”的id。用户拥有的唯一行可能是电子邮件,必须是用户名。
我假设您在javascript中有一个包含所选用户的数组:
var users = new Array("Daniel","Amy","Sandy");
然后你需要使用ajax与php进行通信:
<script>
function ajaxObj( meth, url ) {
var x = new XMLHttpRequest();
x.open( meth, url, true );
x.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
return x;
}
function ajaxReturn(x){
if(x.readyState == 4 && x.status == 200){
return true;
}
}//This can become an external file to link
</script>
所以你可以将数据发布到php:
<script>
var returnedStr = "";
function searchuser(){ //use searchuser function on a button to call
var usersStr = users.toString(); //the string that contain the users separated by ","
var ajax = ajaxObj("POST", "thisurl.php");
ajax.onreadystatechange = function() {
if(ajaxReturn(ajax) == true) {
if(ajax.responseText == "fail"){ //i didn't include this in php, but you can add it yourself if you can't fetch from mysql
echo "Failed";
} else {
returnedStr = ajax.responseText;// when php echos
}
}
}
ajax.send("u="+usersStr);
}
</script>
那么你的php需要处理字符串:
<?php
if(isset($_POST["u"])){
$returnArr = array();
$returnStr = "";
$processedArr = explode(',', $_POST['u']); //Here the posted data will turn into an array
$lengthArr = count($processedArr);
for ($i=0; $i<=$lengthArr; $i++)
{
$req = mysql_query("SELECT id FROM xxx WHERE Username='$processedArr[$i]' LIMIT 1");
while($dnn = mysql_fetch_array($req))
{
array_push($returnArr, $dnn['id']);
}
}
$returnStr = implode(",",$returnArr);
echo ($returnStr);
}
?>
现在Javascript returnedStr
希望是1,2,3
或类似的东西。
如果这不起作用,请发表评论!