Question

我是PHP和HTML编码的新手，所以如果这看起来非常明显，我很抱歉。

我的问题是：一旦用户登录并访问我网站的主页，在他们点击表单上的提交之前，$str_result和$str_comments就会显示。

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;
?>

Answer 1

您实际上应该在if()处写// 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'

之前错过了<a href="salonpage.php?

Answer 2

在脚本的开头，您设置了这个：

$str_result = "<h2>Thanks for your search! Services avaliable are:"

（此外，该行上有一个尾随点，可能应该是;）

然后检查以下内容：

if($str_result == "") $str_result = "<h2>No services found</h2>";

当然不会发生这种情况，因为它永远不会是一个空字符串。

如果您执行此类操作，最后的if语句应该正确触发，您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:"

Answer 3

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>";

    }

您忘了开始{和第二：你应该保留

echo $str_result;
    echo $str_comments;

在}内，只有在点击按钮时才会执行。希望这可以帮助
和

$str_result = "<h2>Thanks for your search! Services avaliable are:" ;

以;
结束

Answer 4

首先纠正这一行你应该在<a href=........>'之前单引号

`$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变量存在时出现。

Answer 5

你应该将代码包含在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，你在<a href="salonpage.php?

之前错过了一个（撇号）

$ str_result在提交表单之前显示

5 个答案: