需要脚本帮助 [英] Script Help required

问题描述

Hiya家伙，


我创建了一个简单的表单，脚本如下：


[HTML]

< form name =" Untitled-11A" method =" post">


型号类型：< select name =？model_id？>

< option value =？model1？/ > Clio

< option value =？model2？/> Espace

< option value =？model3？/> TT

< option value =？model4？/> 306

< option value =？model5？/> Vectra

< / select>


选择颜色：< select name =？colour_id？>

< option value =？colour1？/> Blue

< ;选项值=？colour2？/>绿色

<选项值=？colour3？/>红色

<选项值=？colour4？/>白色

<选项值=？colour5？/>黑色

<选项值=？colour6？/>黄色

< ;选项值=？colour7？/> Silver

< / select>


传输类型：< select name =？transmission_id？>

< option value =？trans1？/>自动

<选项值=？trans2？/>手动

< / select>


位置：< select name =？location_id？ >

< option value =？location1？/> London

< option value =？location2？/>利物浦

< / select>


Car Reg：< input type =" text"名称= QUOT; REG" size =" 20">


< input type =" submit"值= [提交" name =" submit">

< / form>

< / select>

[/ HTML]


然后我有以下脚本，它支持将所选值插入到mysql数据库中，但是你现在最可能已经猜到了，我不知道为什么它有剂量。你们可以查看代码并告诉我我的错误在哪里。


[PHP]


<？php


if（isset（$ _ POST [''reg'']））

$ reg = $ _POST [''reg''];


if（isset（$ _ POST [''model_id'']））

$ mode_id = $ _POST [''model_id''];


if（isset（$ _ POST [''colour_id'']））

$ colour_id = $ _POST [''colour_id''];


if（isset（$ _ POST [''transmisson_id'']））

$ transmisson_id = $ _POST [''transmisson_id''];


if（isset（$ _ POST [''location_id'']））

$ location_id = $ _POST [''location_id''];


//建立与数据库的连接


$ DB = mysql_connect（" host"，" usr"，" pass" ）或死（omg it does work！）;

mysql_select_db（" db"，$ DB）或死（aaah！so close！）;


$ insert =" INSERT INTO car_o2（reg，model_id，colour_id，transmission_id，location_id）VALUES（''$ reg''，''$ model_id''，''$ colour_id'' ，'$ transmission_id''，$ location_id）" ;;


if（mysql_query（$ insert，$ DB）=== FALSE）

{

echo''Car没有插入数据库。< br />'';

}

else

{

echo''Car已成功插入数据库。< br />'';

echo''点击此处返回管理页面' ';

}

？>


[/ PHP]


mysql表有以下列：reg，model_id，colous_ id，transmission_id和location_id。


任何帮助都会受到影响。


keyvan

解决方案

_POST [''reg'']））

reg =

_POST [''reg''] ;


if（isset（

Hiya guys,

I have created a simple form which the script is as follow:

[HTML]
<form name="Untitled-11A" method="post">

MODEL TYPE:<select name=?model_id?>
<option value=?model1?/>Clio
<option value=?model2?/>Espace
<option value=?model3?/>TT
<option value=?model4?/>306
<option value=?model5?/>Vectra
</select>

Choose Colour:<select name=?colour_id?>
<option value=?colour1?/>Blue
<option value=?colour2?/>Green
<option value=?colour3?/>Red
<option value=?colour4?/>White
<option value=?colour5?/>Black
<option value=?colour6?/>Yellow
<option value=?colour7?/>Silver
</select>

Transmission TYPE:<select name=?transmission_id?>
<option value=?trans1?/>Automatic
<option value=?trans2?/>Manual
</select>

Location:<select name=?location_id?>
<option value=?location1?/>London
<option value=?location2?/>Liverpool
</select>

Car Reg:<input type="text" name="reg" size="20">

<input type="submit" value="Submit" name="submit">
</form>
</select>
[/HTML]

Then i have the following script which is suppoe to insert the chosen values into a mysql databse, but as you most probebly have guessed by now, i am not sure why it dosent. can you guys look over the code and enlighten me where my mistake is please.

[PHP]

<?php

if (isset($_POST[''reg'']))
$reg = $_POST[''reg''];

if (isset($_POST[''model_id'']))
$mode_id = $_POST[''model_id''];

if (isset($_POST[''colour_id'']))
$colour_id = $_POST[''colour_id''];

if (isset($_POST[''transmisson_id'']))
$transmisson_id = $_POST[''transmisson_id''];

if (isset($_POST[''location_id'']))
$location_id = $_POST[''location_id''];

//make a connection to database

$DB = mysql_connect("host", "usr", "pass") or die("omg it didnt work!");
mysql_select_db("db", $DB) or die("aaah! so close!");

$insert = "INSERT INTO car_o2 (reg, model_id, colour_id, transmission_id, location_id ) VALUES (''$reg'', ''$model_id'', ''$colour_id'', ''$transmission_id'', $location_id)";

if (mysql_query($insert, $DB) === FALSE)
{
echo ''Car was not inserted into database.<br />'';
}
else
{
echo ''Car was successfully inserted into database.<br />'';
echo ''CLICK HERE to return to admin page'';
}
?>

[/PHP]

The mysql table has the follwoing colums: reg, model_id, colous_id, transmission_id and location_id.

Any help would be appriciated.

keyvan

解决方案

_POST[''reg'']))

reg =

_POST[''reg''];

if (isset(

这篇关于需要脚本帮助的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！