从HTML表单向MySQL数据库添加数据

HI all i've a basic Web Form for putting data into a mysql database, I created code to report if i was connected to my Database correctly and it was so on completion of the form i tested it and it seems to do what i expected but when i goto my database no data was actually entered? I've tried this locally and on a server with both doing the same thing. Here is my two .php forms for you to look that i used on my local machine to test in MAMP just incase i have done something wrong:

你好我一个基本的Web表单将数据到一个mysql数据库,我创建的代码正确地报告如果我是连接到数据库,所以在完成表单我测试它,它似乎做我预期但当我转到我的数据库没有数据实际上是进入吗?我已经在本地和服务器上尝试过了，两者都做同样的事情。这是我的两个。php表单，我在本地机器上测试MAMP，以防我做错了什么:

virtualWalkLog.php

<form action="hazardsform.php" method="POST"  />
  <p>ROUTE: <input type="text" name="ROUTE" /></p>
  <p>ADDRESS: <input type="text" name="ADDRESS" /></p>
  <p>LATITUDE: <input type="text" name="LATITUDE" /></p>
  <p>LONGITUDE: <input type="text" name="LONGITUDE" /></p>
  <p>HAZARD: <input type="text" name="HAZARD" /></p>
  <p>RISK: <input type="text" name="RISK" /></p>
  <input type="submit" value="Submit" />
</form>

hazardsform.php

<?php

define('DB_NAME', 'virtualWalkLog');
define('DB_USER', 'root');
define('DB_PASSWORD', 'root');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
     die('Could not connect: ' . mysql_error());
     }

     $db_selected = mysql_select_db(DB_NAME, $link);

     if (!$db_selected) {
     die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
     }

     $value = $_POST['ROUTE'];
     $value = $_POST['ADDRESS'];
     $value = $_POST['LATITUDE'];
     $value = $_POST['LONGITUTE'];
     $value = $_POST['HAZARD'];
     $value = $_POST['RISK'];

     $sql = "INSERT INTO rmbhazards (ROUTE, ADDRESS, LATITUDE, LONGITUDE, HAZARD, RISK) VALUES ('$value', '$value2', 
     '$value3', '$value4', '$value5', '$value6')";

     mysql_close();

Many Thanks in advance

提前感谢

6 个解决方案

#1

Going through your script quickly you need to call mysql_query($sql) after

快速遍历脚本之后，需要调用mysql_query($sql)

$sql = "INSERT INTO rmbhazards (ROUTE, ADDRESS, LATITUDE, LONGITUDE, HAZARD, RISK) VALUES ('$value', '$value2', '$value3', '$value4', '$value5', '$value6')";

$sql =“插入元危险(路由、地址、纬度、经度、危害、风险)值(‘$value’、‘$value2’、‘$value3’、‘$value4’、‘$value5’、‘$value6’)”;

mysql_sql query will actually execute the query.

mysql_sql查询将实际执行查询。

Also as $value should be unique

同样，$value应该是唯一的

 $value = $_POST['ROUTE'];
 $value2 = $_POST['ADDRESS'];
 $value3 = $_POST['LATITUDE'];

 -----

SUGGESTION Since you have just begin ..I will suggest you try mysql_* for just concepts but use mysqli_* or PDO .. You shold also know about sql injection

你刚开始提出建议。我建议您尝试使用mysql_*，但使用mysqli_*或PDO。您还应该了解sql注入

Here are some tutorials to help you

这里有一些教程可以帮助您

http://php.net/manual/en/security.database.sql-injection.php

http://php.net/manual/en/book.pdo.php

http://php.net/manual/en/book.mysqli.php

#2

you are not exectuing your query, this is why no data is inserted. Try to place after

您没有执行查询，这就是为什么没有插入数据。试着在

$sql = "INSERT INTO rmbhazards (ROUTE, ADDRESS, LATITUDE, LONGITUDE, HAZARD, RISK) VALUES ('$value', '$value2', '$value3', '$value4', '$value5', '$value6')";

this

这

$result = mysql_query($sql);

also all values are in one variable $value, so you will end up with all the same result in your table so change to this to fit your query

而且所有的值都在一个变量$value中，所以您的表将得到所有相同的结果，因此更改为适合您的查询

 $value = $_POST['ROUTE'];
 $value2 = $_POST['ADDRESS'];
 $value3 = $_POST['LATITUDE'];
 $value4 = $_POST['LONGITUTE'];
 $value5 = $_POST['HAZARD'];
 $value6 = $_POST['RISK'];

I would also sugeest you to stop using mysql_ api since they are depecrated, please switch to PDO or mysqli

我还想让您停止使用mysql_ api，因为它们是depecare，请切换到PDO或mysqli。

Furthermore you are ready to mysql injection. there is a nice tutorial here which explain you everything about that -> How can I prevent SQL injection in PHP?

此外，您已经准备好进行mysql注入。这里有一个很好的教程，可以向您解释关于这个的一切——>如何防止PHP中的SQL注入?

#3

you are assigning values to only one variable $value here

这里只为一个变量赋值$value

 $value = $_POST['ROUTE'];
 $value = $_POST['ADDRESS'];
 $value = $_POST['LATITUDE'];
 $value = $_POST['LONGITUTE'];
 $value = $_POST['HAZARD'];
 $value = $_POST['RISK'];

should be

应该是

 $value = $_POST['ROUTE'];
 $value2 = $_POST['ADDRESS'];
 $value3 = $_POST['LATITUDE'];
 $value4 = $_POST['LONGITUTE'];
 $value5 = $_POST['HAZARD'];
 $value6 = $_POST['RISK'];

Also call mysql_query($sql); for running the query.

也叫mysql_query($ sql);运行查询。

#4

just rename:

重命名:

$value = $_POST['ROUTE'];
$value2 = $_POST['ADDRESS'];
$value3 = $_POST['LATITUDE'];
$value4 = $_POST['LONGITUTE'];
$value5 = $_POST['HAZARD'];
$value6 = $_POST['RISK'];

#5

You are capturing all the input fields value into one variable. You need to execute mysql_query for it to work. Change this :-

将所有输入字段值捕获到一个变量中。您需要执行mysql_query以使其工作。改变这个:

     $value = $_POST['ROUTE'];
     $value = $_POST['ADDRESS'];
     $value = $_POST['LATITUDE'];
     $value = $_POST['LONGITUTE'];
     $value = $_POST['HAZARD'];
     $value = $_POST['RISK'];

to:-

:- - -

 $value = $_POST['ROUTE'];
 $value2 = $_POST['ADDRESS'];
 $value3 = $_POST['LATITUDE'];
 $value4 = $_POST['LONGITUTE'];
 $value5 = $_POST['HAZARD'];
 $value6 = $_POST['RISK'];

Once you have done that, you need to call mysql_query($sql) to execute the query.

完成之后，需要调用mysql_query($sql)来执行查询。

#6

You kep all the variables as the same name

你把所有的变量都记为同一个名字

 $value = $_POST['ROUTE'];
 $value = $_POST['ADDRESS'];
 $value = $_POST['LATITUDE'];
 $value = $_POST['LONGITUTE'];
 $value = $_POST['HAZARD'];
 $value = $_POST['RISK'];

change them to unique id's (as you referenced in the sql statement)

将它们更改为唯一id(如您在sql语句中引用的那样)

 $value1 = $_POST['ROUTE'];
 $value2 = $_POST['ADDRESS'];
 $value3 = $_POST['LATITUDE'];
 $value4 = $_POST['LONGITUTE'];
 $value5 = $_POST['HAZARD'];
 $value6 = $_POST['RISK'];

and change your query statement to actually execute

并更改查询语句以实际执行。

 $result = mysql_query("INSERT INTO rmbhazards (ROUTE, ADDRESS, LATITUDE, LONGITUDE, HAZARD, RISK) VALUES ('$value', '$value2', 
 '$value3', '$value4', '$value5', '$value6')");

#1