来自数据库的数据没有显示，HELP？

Question

如何从我的数据库中获取数据以显示。我对PHP或MySQL不是很有经验。

我没有收到错误消息，但没有数据显示，所以我做错了什么？

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 
} 
} 
?> 

谢谢！

Answer 1

确保您正在执行的查询返回记录。你可以通过添加一个echo语句来检查它，它将在你的屏幕上打印查询。复制并运行它againist的database.You可以使用任何MySQL的前端工具（PHP myadmin，mysqlyog运行查询，如果有查询任何错误，你可以看到，那么，

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

$result = mysql_query ($query); 

Answer 2

它应该是这样的：

<?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是为了防止MySQL的注入这是一个严重的风险