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我是新來的php,我想從多個表搜索 其關於搜索。數據應以表格形式顯示。我嘗試了很多,但無法解決問題。我看了很多教程,與許多同事見面,但找不到任何可能的解決方案。請幫幫我。我會成爲你的一個慷慨行爲。如何在多個表中搜索php mysql並在表中顯示結果
我有4個表
(1)user
its have user_name,id
(2)message
its have user_name (FK),messages
(3)
images
user_name(FK),images
(4)
videos
user_name,videos
我想,當我將在搜索框中填寫用戶名,如用戶「蛙」
它應該顯示在表中的數據
蛙有這個圖片,視頻,消息
以表格的形式
這是我的形式
/*search.php*/
<form action="join.php" method="post" class="navbar-form navbar-right">
<div class="input-group">
<input type="Search" name="user_name" value="" placeholder="Search..." class="form-control" />
<div class="input-group-btn">
<button class="btn btn-info">
<span class="glyphicon glyphicon-search"></span>
</button>
</div>
</div>
<a href="logout.php" class="btn btn-danger">logout <span class="glyphicon glyphicon-log-out"></span></a>
</form>
和我的搜索
/*join.php*/
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "terror_combat";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "select *from user a
join messages b on a.user_name = b.user_name
join images c on a.user_name = c.user_name
join videos d on a.user_name = d.user_name";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "User Name: " . $row["user_name"]. "<br />";
echo "Message: " . $row["message"]. "<br />";
echo "Video Path: " . $row["name"]. "<br />";
echo'<img height="300" width="300" src="data:image;base64,'.$rows['image_path'].'">';
//you get more data as your wise................
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
我使用PHP PHP和我的SQL和XAMP
我想,當我會寫在搜索框中的用戶名like用戶「rana」
它應該在表中顯示數據
rana has this ima GES,視頻,消息
在感謝表格形式求助