來自數據庫的數據沒有顯示，HELP？

Question

如何從我的數據庫中獲取數據以顯示。我對PHP或MySQL不是很有經驗。

我沒有收到錯誤消息，但沒有數據顯示，所以我做錯了什麼？

PHP

<?php 
if(strlen(trim($_POST['search'])) > 0) { 

$search = "%" . $_POST["search"] . "%"; 
$searchterm = "%" . $_POST["searchterm"] . "%"; 

mysql_connect ("cust-mysql-123-03", "", ""); 
mysql_select_db ("weezycouk_641290_db1"); 
if (!empty($_POST["search_string"])) 
{ 

} 

$query = "SELECT name,lastname,email FROM contact WHERE name LIKE '%$search%' AND 
lastname  LIKE '%$searchterm%'"; 

$result = mysql_query ($query); 
echo mysql_error(); 
if ($result) { 
while ($row = mysql_fetch_assoc($result)) { 

echo $row["name"]; 
echo $row["lastname"]; 
echo $row["email"]; 
} ?> 


<?php echo $row["name"]; ?> 
<br> 
<?php echo $row["lastname"]; ?> 
<br> 
<?php echo $row["email"]; ?> 


<?php 
} 
} 
?> 

謝謝！

Answer 1

確保您正在執行的查詢返回記錄。你可以通過添加一個echo語句來檢查它，它將在你的屏幕上打印查詢。複製並運行它againist的database.You可以使用任何MySQL的前端工具（PHP myadmin，mysqlyog運行查詢，如果有查詢任何錯誤，你可以看到，那麼，

$query = "SELECT name,lastname,email FROM contact WHERE name LIKE '%$search%' AND 
lastname  LIKE '%$searchterm%'"; 
//the below line will print the query on the screen 
echo $query; 

$result = mysql_query ($query); 

Answer 2

它應該是這樣的：

<?php 
if(strlen(trim($_POST['search'])) > 0) { 

mysql_connect ("cust-mysql-123-03", "", ""); 
mysql_select_db ("weezycouk_641290_db1"); 

$query = "SELECT name,lastname,email FROM contact WHERE name LIKE '%" . mysql_real_escape_string($_POST['search']) . "%' AND lastname LIKE '%" . mysql_real_escape_string($_POST['searchstring']) . "%'"; 

$result = mysql_query ($query); 
echo mysql_error(); 
if ($result) { 
while ($row = mysql_fetch_assoc($result)) { 

echo $row["name"]; 
echo $row["lastname"]; 
echo $row["email"]; 
} ?> 


<?php echo $row["name"]; ?> 
<br> 
<?php echo $row["lastname"]; ?> 
<br> 
<?php echo $row["email"]; ?> 


<?php 
} 
} 
?> 

的mysql_real_escape_string是爲了防止MySQL的注入這是一個嚴重的風險