我是php新手,正在创建一个社交网站,人们可以登录注册,我希望人们互相搜索,当结果出来时,用户可以点击结果并转到搜索用户
这是我的代码search.php
<?php
if (isset($_POST['submit-search'])) {
$search = mysqli_real_escape_string($con,$_POST['search']);
$sql = "SELECT * FROM users WHERE fullname LIKE '%$search%' OR username LIKE '%$search%' OR email LIKE '%$search%' ";
$result = mysqli_query($con,$sql);
$queryResults = mysqli_num_rows($result);
if ($queryResults > 0) {
while ($row = mysqli_fetch_assoc($result)) {
echo "<a href='profile.php?id=".$row['id']."'><div>
<h3>".$row['fullname']."</h3>
</div></a>";
}
}else{
echo "no results";
}
}
?>
profile.php中的代码
<?php
if(isset($_GET['user'])){
$id = $_GET['user'];
$query = "SELECT * FROM users WHERE email=$email";
$fire = mysqli_query($con,$query) or die("Can not fetch the data.".mysqli_error($con));
$user = mysqli_fetch_assoc($fire);
}
?>
实际上每当我点击搜索结果时,我都能够获得该配置文件的ID但没有输出页面变为空白plzz帮我这个狗屎
发布于 2019-02-21 09:42:23
首先,如果你得到$ id,你应该选择id而不是电子邮件。并且您没有获得id,但用户不存在。你应该这样试试。
<?php
if(isset($_GET['id'])){
$id = $_GET['id'];
$query = "SELECT * FROM users WHERE id=$id";
$fire = mysqli_query($con,$query) or die("Can not fetch the data.".mysqli_error($con));
$user = mysqli_fetch_assoc($fire);
}
?>
https://stackoverflow.com/questions/-100003118
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