Question

我有以下表格，允许用户选择酒店预订的日期/房间。

<form action="booking-form.php" method="post">
  <fieldset>
    <div class="select-date">
      <label>Arrival Date</label>
      <select name="arrd" class="day">
        <option value="1" >1</option>
        ... snip ...
        <option value="24" selected="selected">24</option>
      </select>

      <select name="arrm" class="month">
        <option value="01" >January</option>
        ... snip ...
        <option value="11" selected="selected">November</option>
      </select>
      <select name="arry" class="year">
        <option value="2009" selected="selected">2009</option>
        ... snip ...
      </select>
    </div>
    <div class="more">
    <div class="info">
      <label>Days</label>
      <select name="days">
        <option value="1">01</option>
        ... snip ...
        <option value="7" selected="selected">07</option>
      </select>
    </div>
    <div class="info">
      <label>Rooms</label>
      <select name="rooms">
        <option value="1">01</option>
        ... snip ...
      </select>
    </div>
    </div>
    <input type="submit" value="Check availabilty"/>
    <input type="hidden"  name="submitted" value="TRUE"/>
  </fieldset>
</form>

这是应该处理从表单传递的变量并通过url传递给预订提供者的php。当表单只提交一个没有错误的白色屏幕时，没有任何反应。

<?php

 ini_set ('display_errors',1);  
 error_reporting (E_ALL & ~ E_NOTICE);  

 if (isset($_POST['submitted'])){

  $errors = array();

  // A bunch of if's for all the fields and the error messages.


  if (empty($_POST['arrd'])) {
    $errors[] = 'Please select a day';
} else{
    $arrd = ($_POST['arrd']);
}

if (empty($_POST['arrm'])) {
    $errors[] = 'Please select a month';
} else{
    $arrm = ($_POST['arrm']);
}

if (empty($_POST['arry'])) {
    $errors[] = 'Please select a year';
} else{
    $arry = ($_POST['arry']);
}

if (empty($_POST['non'])) {
    $errors[] = 'Please choose a number of nights';
} else{
    $non = ($_POST['non']);
}


if (empty($_POST['norooms'])) {
    $errors[] = 'Please choose a number of rooms';
} else{
    $norooms = ($_POST['norooms']);
}


if (empty($errors)){
$arrdate = $arrd . " " . $arrm ."" . $arry; 

$url ="https://portals.uk.rezlynx.net/DHPPORTAL/wfrmpakquery.aspx?SiteID=DHOLDHALL&arrdate=$arrdate&non=$non&norooms=$norooms";
 header('Location: $url');
exit();

 }

 else{
 foreach($errors as $msg){

 echo"-msg<br/>\n";
 }

 }

 ?>

有人能看到我错过的东西吗？

提前致谢..

丹

Answer 1

该行

header('Location: $url');

应该是

header("Location: $url");

不会在单引号内评估变量。

Answer 2

如果你通过PHP的lint checker（php -l）运行你的代码，你会发现第62行（最后一行）有语法错误

你只是缺少需要关闭初始“if”语句的大括号。

Answer 3

问题可能是

header('Location: $url');
exit();

尝试

header('Location: '.$url);
exit();

或

header("Location: $url");
exit();
# using the double quotes will parse the $url variable

Answer 4

您是否只能在表单的操作属性中使用该网址并将方法设置为GET？

<form action="https://portals.uk.rezlynx.net/DHPPORTAL/wfrmpakquery.aspx?SiteID=DHOLDHALL" method="get">
    [...]
</form>

使用php将表单变量传递到url

4 个答案: