php 如何获取 HTML 表单以从 Mysql 数据库中查询和生成结果

How to get a HTML Form to query and produce results from Mysql datebase

提问by user2013520

I have created a form that has one text field and one select box, I also have created a button. Now what I want to happen is when the button is pressed I want the string in the forms to be matched up to my database and then I want the results of those matches to be returned and shown on the screen.

我创建了一个包含一个文本字段和一个选择框的表单，我还创建了一个按钮。现在我想要发生的是，当按下按钮时，我希望表单中的字符串与我的数据库相匹配，然后我希望返回这些匹配的结果并显示在屏幕上。

For example if I was to type 'History' into my text field box and I also selected 'level 2' from my select box when the button (Submit Button) was pressed I would want returned on the page everything that matches up with the word history and the selection of level 2 in my database.

例如，如果我要在文本字段框中键入“历史记录”，并且在按下按钮（提交按钮）时我还从选择框中选择了“级别 2”，我希望在页面上返回与该词匹配的所有内容我的数据库中的历史记录和级别 2 的选择。

I know that I have to connect my page with my Database through PHP and I have successfully done that, but what I don't understand is how to then get my HTML Form to Query the database and provide results back onto the screen

我知道我必须通过 PHP 将我的页面与我的数据库连接，我已经成功地做到了这一点，但我不明白的是如何让我的 HTML 表单查询数据库并将结果返回到屏幕上

In terms of a example website that is very similar to the concept I would like to create take a look at this webpage. http://search.ucas.com/cgi-bin/hsrun/search/search/search.hjx;start=search.HsSearch.run?y=2013&w=H(UCAS Course Search) this website has multiple text fields and select boxes and a submit button exactly as I am trying to create and the results provided are only the results that match with what has been searched.

就一个与我想创建的概念非常相似的示例网站而言，请查看此网页。http://search.ucas.com/cgi-bin/hsrun/search/search/search.hjx;start=search.HsSearch.run?y=2013&w=H(UCAS Course Search) 这个网站有多个文本字段，选择框和提交按钮与我试图创建的完全一样，提供的结果只是与搜索结果相匹配的结果。

The code below works in terms of it links my text-field to my database but I can't get my text-field and my select box to link and query the database, only one or the other. I want them both to work from the same button (search button)

下面的代码将我的文本字段链接到我的数据库，但我无法让我的文本字段和我的选择框链接和查询数据库，只有一个或另一个。我希望它们都使用同一个按钮（搜索按钮）

<form  method="post" action="Webpage here"  id="searchform">
  <input  type="text" name="name">
  <input  type="submit" name="submit" value="Search">
</form>
<form id="form1" name="ExamBoard" method="post" action="Webpage here">
  <label for="select"></label>
  <select name="ExamBoard" id="select">
    <option value="EB1" selected="selected">EB1</option>
    <option value="EB2">EB2</option>
    <option value="EB3">EB3</option>
  </select>
  <input  type="submit" name="submit" value="Search">
</form>
<p>&nbsp;</p>
<p>
<?php
 if(isset($_POST['submit'])){
 if(isset($_GET['go'])){
 if(preg_match("/^[  a-zA-Z]+/", $_POST['name'])){
 $name=$_POST['name'];
  //connect  to the database
 $db=mysql_connect  ("Name", "User",  "Password*") or die ('I cannot connect to the      database  because: ' . mysql_error());
 //-select  the database to use
 $mydb=mysql_select_db("Table Name");
  //-query  the database table
  $sql="SELECT  ID, CourseName, ExamBoard FROM subjects WHERE CourseName LIKE '%" . $name .  "%' ";
   //-run  the query against the mysql query function
   $result=mysql_query($sql);
    //-create  while loop and loop through result set
    while($row=mysql_fetch_array($result)){
      $CourseName  =$row['CourseName'];
      $ID=$row['ID'];
      $ExamBoard=$row['ExamBoard'];
  //-display the result of the array
  echo "<ul>\n";
  echo "<li>" . "<a  href=\"search.php?id=$ID\">"   .$CourseName . " " .  "</a></li>\n" ;
  echo  $ExamBoard . " " .  "</a>\n";
  echo "</ul>";
  }
  }
  }
  }
  ?>

回答by user1551869

New HTML File

新建 HTML 文件

newHTML.htm

新HTML.htm

<form  method="post" action="Webpage here"  id="searchform">
  <input  type="text" name="name">
  <input  type="submit" name="submit" value="Search">
</form

New PHP file

新建 PHP 文件

newPHP.pfp

新建PHP.pfp

<?php
 if(preg_match("/^[  a-zA-Z]+/", $_REQUEST['name'])){
 $name=$_REQUEST['name'];
  //connect  to the database
 $db=mysql_connect  ("Name", "User",  "Password*") or die ('I cannot connect to the      database  because: ' . mysql_error());
 //-select  the database to use
 $mydb=mysql_select_db("Table Name");
  //-query  the database table
  $sql="SELECT  ID, CourseName, ExamBoard FROM subjects WHERE CourseName LIKE '%" . $name .  "%' ";
   //-run  the query against the mysql query function
   $result=mysql_query($sql);
    //-create  while loop and loop through result set
    while($row=mysql_fetch_array($result)){
      $CourseName  =$row['CourseName'];
      $ID=$row['ID'];
      $ExamBoard=$row['ExamBoard'];
  //-display the result of the array
  echo "<ul>\n";
  echo "<li>" . "<a  href=\"search.php?id=$ID\">"   .$CourseName . " " .  "</a></li>\n" ;
  echo  $ExamBoard . " " .  "</a>\n";
  echo "</ul>";
}
?>

回答by user1551869

You have to ask yourself Do you want to do this with AJAX or with normal screen refreshes..

你必须问问自己，你想用 AJAX 还是普通的屏幕刷新来做到这一点。

With normal screen refreshes (The easy way)

使用正常的屏幕刷新（简单的方法）

Take your HTML and place it in one file (HTMLfile.html) Take your PHP and place it in another file (PHPCode.php)

将您的 HTML 放入一个文件 (HTMLfile.html) 将您的 PHP 放入另一个文件 (PHPCode.php)

HTML file form calls the PHP script from action attribute.

HTML 文件形式从 action 属性调用 PHP 脚本。

PHP file renders the response.

PHP 文件呈现响应。

Dont mash up the view and the controlling application.

不要混搭视图和控制应用程序。

There is other ways to do this in one script if you want to use AJAX.

如果您想使用 AJAX，还有其他方法可以在一个脚本中执行此操作。