我是PHP和HTML编码的新手，所以如果这看起来很明显，我很抱歉。
我的问题是：一旦用户登录并转到我的网站主页，在他们单击表单上的提交之前，将显示$str_result和$str_comments。
这是我的代码：
<?php
//If connected to the database get services names from database and write out DropDownMenu
mysqli_select_db($db_server, $db_database);
$query = "SELECT ID, Name FROM categories ORDER BY Name";
$result = mysqli_query($db_server, $query);
if (!$result) die("Query failed: " . mysqli_error($db_server));
while($row = mysqli_fetch_array($result)){
$str_options .= "<option value= '" . $row[ 'ID'] . "'>";
$str_options .= $row['Name'];
$str_options .= "</option>";
}
mysqli_free_result($result);
// Your code here to handle a successful verification
$str_result = "<h2>Thanks for your search! Services avaliable are:" .
$category = clean_string($db_server, $_POST["categories"]) . "</h2>";
?>
<!--form-->
<form method="post" action="nihome.php"><p>I am searching for</p>
<select name="categories"><?php echo $str_options; ?></select>
<br />
<input type="submit" id="submit" name="submit" value="Submit" />
</form>
<?php
//Capture form data, if anything was submitted
if (isset($_POST['categories']) and ($_POST['categories'] != '')){
$category = clean_string($db_server, $_POST['categories']);
// create the SQL query
$query = "SELECT salon.ID AS ID, categories.Name as Category, salon.salon_name AS Salon, services.name AS Service, servicesoffered.price AS price FROM services
JOIN categories ON services.cID = categories.ID
JOIN servicesoffered ON servicesoffered.serviceID = services.ID
JOIN salon ON servicesoffered.salonID = salon.ID WHERE categories.ID=$category";
// query the database
mysqli_select_db($db_server, $db_database);
$result = mysqli_query($db_server, $query);
if (!$result) die("Database access failed: " . mysqli_error($db_server));
// if there are any rows, print out the contents
while ($row = mysqli_fetch_array($result)) {
$str_result .= '<h3>' . $row['Salon'] . ',</h3><p>' .
$row['Service'] . ", £" .
$row['price'] .'</p>' .<a href="salonpage.php?salonid=' . $row['ID'] .'">Click here to view or add to salon reviews</a>';
}
if($str_result == "") $str_result = "<h2>No services found</h2>";
} else {
$str_result = '<h2>No service was requested</h2>';
}
mysqli_close($db_server);
echo $str_result;
echo $str_comments;
?>
你实际上应该写一个if()，上面写着// Your code here to handle a successful verification：
if (!empty($_POST) /* && all your other validation routines */)
{
$str_result = "<h2>Thanks for your search! Services avaliable are:" .
$category = clean_string($db_server, $_POST["categories"]) . "</h2>";
}
此外，您在<a href="salonpage.php?之前错过了'
在脚本的开头，设置以下内容：
$str_result = "<h2>Thanks for your search! Services avaliable are:"
(另外，那条线上有一个尾随的点，可能应该是一个;(
然后检查以下内容：
if($str_result == "") $str_result = "<h2>No services found</h2>";
当然，这永远不会发生，因为它从一开始就不是一个空字符串。
如果你做这样的事情，你的if语句在最后应该正确触发，你应该得到预期的结果：
$str_result = "";
[ ... ]
if (isset($_POST['categories']) and ($_POST['categories'] != '')){
// Set it inside this if statement.
// So it will stay empty when nothing was submitted
$str_result = "<h2>Thanks for your search! Services avaliable are:"
首先更正此行您应该在<a href=........>'之前加上单引号
`$row['price'] .'</p>' .'<a href="salonpage.php?salonid=' . $row['ID'] .'">Click here to view or add to salon reviews</a>';
和将$str_result和$str_comments封装在isset中，然后它们只会在categories变量存在时才出现。
你应该试试这个
if($str_result == "")
{ $str_result = "<h2>No services found</h2>";
} else {
$str_result = '<h2>No service was requested</h2>';
}
mysqli_close($db_server);
echo $str_result;
echo $str_comments;
}
这里有两件事非常重要。
第一：
if($str_result == "")
{ $str_result = "<h2>No services found</h2>";
}
你忘记了起始{和
第二：
你应该保持
echo $str_result;
echo $str_comments;
在}里面，这将使它仅在单击按钮时执行。希望这有帮助
和
$str_result = "<h2>Thanks for your search! Services avaliable are:" ;
以;结束
您应该将代码包含在 if 条件中。
即
if (!empty($_POST){
if($str_result == "") $str_result = "<h2>No services found</h2>";
} else {
$str_result = '<h2>No service was requested</h2>';
}
}
另外，您在<a href="salonpage.php?之前缺少一个(撇号('