所以,我正在做的是使用AJAX搜索存储在数据库中的记录,但是当它打印回调数据时,第一个记录是重复的:
我的代码:
$gUser = $_GET['q'];
$connect = mysql_connect("localhost", "root", "") or die("Could not connect to the server");
mysql_select_db("socialj") or die("Could not connect to the database");
$result = mysql_query("SELECT fullname, email FROM users WHERE fullname LIKE '%$gUser%' ");
while($array[] = mysql_fetch_array ($result))
{
foreach($array as $r)
{
echo $r['fullname'].' | '.$r['email'].'<br>';
}
}
&GT;
JavaScript代码:
$(document).ready(function (){
$('#searchh').on('submit', function (e){
e.preventDefault();
var sVal = $('#search').val();
$.ajax({
type: 'get',
url: 'profile.php',
data: {q : sVal},
success: function (data) {
alert(data);
}
});
});
});
你能帮助我吗?我不知道发生了什么......谢谢。答案 0 :(得分:0)
替换此
while($array[] = mysql_fetch_array($result)) {
foreach($array as $r) {
echo $r['fullname'] . ' | ' . $r['email'] . '<br>';
}
}
用这个
while($row = mysql_fetch_array($result)) {
echo $row['fullname'] . ' | ' . $row['email'] . '<br>';
}