0

哎呀对不起，我忘记改了。在此代码中，它将显示 cakename 列表。

<?php
$sql="SELECT * FROM $tbl_name ORDER BY c_id ASC";
$result=mysql_query($sql);

while($rows=mysql_fetch_array($result)){
$c_id=$rows['c_id']; // idnumber of cake
$cakename=$rows['cakename']; //this is the name of cake
$options.="<OPTION VALUE=\"$c_id\">$cakename</OPTION>"; //it will list all the cake
// close while loop 
}
?>

<select name="order_id">
<option value=0>Select Cake:<?php echo $options; ?>
</select>

而这个，当用户选择了 cake1.it 将显示细节，价格和图片。这是我的问题。我想结合所有的代码。我不知道如何显示图片。

<html>
<form action="current_file.php" method="POST">
<select name="order_id" id="order_id" onChange="this.form.submit()">
<option value="">Select Cake:</options>
<option value="1">cake1</options>
<option value="2">cake2</options>
<option value="3">cake3</options>
</select>
</form>

<?php

$order_id=$_POST['order_id']; 

if (!empty($order_id)){

   $sql="SELECT * FROM $tbl_name WHERE c_id = $order_id";
    $result=mysql_query($sql);
    while($rows=mysql_fetch_array($result)){
        $price=$rows['price'];
        $details=$rows['details'];
        $picture=$rows['picture'];
       // close while loop 
     }
     echo $details;
echo $price;
}//close if statement

?>

是的，所有的代码都在工作。但我只想成为一个。明白了吗？像这样但不工作：

<?php
  $order_id=$_POST['order_id']; 

    if (!empty($order_id)){
    $sql="SELECT * FROM $tbl_name WHERE c_id = $order_id";
    $result=mysql_query($sql);

    while($rows=mysql_fetch_array($result)){

    $price=$rows['price'];
    $details=$rows['details'];
    $picture=$rows['picture'];
    $c_id=$rows['c_id']; // idnumber of cake
    $cakename=$rows['cakename']; //this is the name of cake
    $options.="<OPTION VALUE=\"$c_id\">$cakename</OPTION>"; //it will list all the cake
    // close while loop 
    }
    ?>

    <form action="current_file.php" method="POST">
    <select name="order_id" id="order_id" onChange="this.form.submit()">
    <option value=0>Select Cake:<?php echo $options; ?>
    </select>
<?php
    echo $details;
    echo $price;
    echo $picture; 
?>
</form>
4

2 回答 2

0 
while($rows=mysql_fetch_array($result)){
    $c_id=$rows['c_id'];
    $options.="<OPTION VALUE=\"$c_id\">$cakename</OPTION>";
    // close while loop 
}
?>

<select name="order_id">
    <option value=0>Select Cake:<?php echo $options; ?>
</select>

您在此语句中有 html 错误，因为您的 html 将是这样的

<option value=0>Select Cake:<option value='1'>what ever</option>

所以你必须关闭第一个选择选项，所以它应该是：

while($rows=mysql_fetch_array($result)){
    $c_id=$rows['c_id'];
    $options.="<OPTION VALUE=\"$c_id\">$cakename</OPTION>";
    // close while loop 
}
?>

<select name="order_id">
    <option value=0>Select Cake:</option>
    <?php echo $options; ?>
</select>
于 2013-03-08T17:54:31.110 回答
0

试试这个：

现在只有一个 PHP 文件会检查帖子消息。如果$_POST['pet_id']存在，它将显示项目详细信息，否则它将向用户显示一个选择框。

<?php
    if (!empty($_POST['order_id'])) {
        $order_id = $_POST['order_id'];
        $sql = "SELECT details FROM $tbl_name WHERE c_id = $order_id";
        $result = mysql_query($sql);
        while ($rows = mysql_fetch_array($result)) {

            // $c_id=$rows['c_id'];
            $details = $rows['details'];
            $price = $rows['price'];
            $picname=$rows['picname']; // this is the name of the picture
            $picture = '<img src="/images/' . $picname . '.jpg" />'; //i change the $pet_id to $picname the pet_id are not inluded 
        } // close while loop 
        echo $details;
        echo $price; // What's the price? You haven't selected price from db...
        echo $picture;
    } else {
        $sql = "SELECT * FROM $tbl_name ORDER BY c_id ASC";
        $result = mysql_query($sql);
        $options = ''; // First Create an empty variable then fill it in a loop
        while ($rows = mysql_fetch_array($result)) {
            $c_id = $rows['c_id'];
            $cakename = $rows['c_cakename'];
            $options .= "<option value=\"$c_id\">$cakename</option>\n";
        }
?>
    <!-- Create a form to send data -->
    <form action="yourtarget.php" method="post">
        <select name="order_id" onChange="this.form.submit()">
            <!-- You are putting all of your options inside the first option tag. Make sure you close the first one and then echo rest of options. -->
            <option value=0>Select Cake:</option>
            <?php echo $options; ?>
        </select>
    </form>
<?php } ?>

我对图像的想法是通过它们的 item-id 将它们存储在一个文件夹中，您可以将图像名称或路径存储在 db 中。有太多方法可以做到这一点。

于 2013-03-08T17:57:56.990 回答