Question

这是我的代码，我想要做的是在另一个页面中显示我的朋友的名字作为超链接，当点击链接时，我想显示数据。

<?php
$con=mysqli_connect("localhost","root","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM friends");



while($row = mysqli_fetch_array($result))
{

echo "<p>" . $row['FirstName'] . "</p>";
echo "<p>" . $row['LastName'] . "</p>";
echo "<p>" . $row['Email'] . "</p>";
echo "<p>" . $row['Address'] . "</p>";
}

mysqli_close($con);
?>

Answer 1

<?php
$con=mysqli_connect("localhost","root","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM friends");



while($row = mysqli_fetch_array($result))
{?>php

<a href=<?php "echo "<p>" . $row['FirstName'] . "</p>";"?>> `your page name`</a><?php
echo "<p>" . $row['LastName'] . "</p>";
echo "<p>" . $row['Email'] . "</p>";
echo "<p>" . $row['Address'] . "</p>";
}

mysqli_close($con);
?>

Answer 2

您的问题文件发生了变化：

while($row = mysqli_fetch_array($result))
{

echo "<p>" . $row['FirstName'] . "</p>";
echo "<p>" . $row['LastName'] . "</p>";
echo "<p>" . $row['Email'] . "</p>";
echo "<p>" . $row['Address'] . "</p>";
}

到

while($row = mysqli_fetch_array($result))
{
echo '<a href="show.php?UserId='.$row['UserId'].'"> . $row['FirstName'] . "</a>";
}

show.php

$con=mysqli_connect("localhost","root","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_GET['UserId'])){
if(is_numeric($_GET['UserId'])){
$UserId = (int)$_GET['UserId'];

$result = mysqli_query($con,"SELECT * FROM friends WHERE UserId= $UserId");
if(count($result)){
while($row = mysqli_fetch_array($result))
{

echo "<p>" . $row['FirstName'] . "</p>";
echo "<p>" . $row['LastName'] . "</p>";
echo "<p>" . $row['Email'] . "</p>";
echo "<p>" . $row['Address'] . "</p>";
}}}}

重要提示：我不确定什么是UserId列名，因此请进行更改。

另外，请检查语法错误。

Answer 3

尝试使用数据库中的增量ID连接它。因此，您朋友表中的每一行都必须具有唯一的ID。通过这种方式，您可以连接它并从您的朋友表中获取行数据。

所以你的朋友表必须如下所示：

friend_id (unique id)
firstname
lastname
email
address

您的代码可能类似（这不是任何攻击的保存）：

<?php
$con = mysqli_connect("localhost","root","abc123","my_db");
if (mysqli_connect_errno()) {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

if(isset($_GET['friendid'])) { 
    $result = mysqli_query($con,"SELECT * FROM friends");
    while($row = mysqli_fetch_array($result)) {
       echo "<p>" . $row['FirstName'] . "</p>";
       echo "<p>" . $row['LastName'] . "</p>";
       echo "<p>" . $row['Email'] . "</p>";
       echo "<p>" . $row['Address'] . "</p>";
       echo "<p><a href=\"_YOUR_SCRIPT_URL_HERE_?friendid=".$row['friend_id']."\" target=\"_self\">Details</a></p>";
    }
} else {
   $result = mysqli_query($con,"SELECT * FROM friends WHERE friend_id = '".$_GET['friendid']."'");
   $row = mysqli_fetch_assoc($result);  
   echo "<p>" . $row['FirstName'] . "</p>";
   echo "<p>" . $row['LastName'] . "</p>";
   echo "<p>" . $row['Email'] . "</p>";
   echo "<p>" . $row['Address'] . "</p>";  
} 
mysqli_close($con);
?>

从MySql数据库中搜索并将结果显示为超链接

3 个答案: