我正在做一个项目，在AddJob页面中，我需要根据用户选择rocord保存位置的含义在myslq中的表中添加记录。
这是我的代码，直到现在，它只将记录保存到Food Table中。我可以将其设为可选项。
<?php
session_start();
if( isset($_SESSION['username']) ){
if(isset($_POST['add'])){
if(isset($_POST['add']))
{
$errorMessage = "";
if(($_POST['lists'])== 0) // trying to get error if user don't choose.
{
$errorMessage .= "<li>You Forgot to Choose !</li>";
}
if(empty($_POST['JobName']))
{
$errorMessage .= "<li>You Forgot To Enter A Job Name !</li>";
}
if(empty($_POST['Description']))
{
$errorMessage .= "<li>You Forgot To Enter A Description !</li>";
}
if(empty($_POST['NoStudent']))
{
$errorMessage .= "<li>You Forgot To Enter A Student Number !</li>";
}
if(empty($_POST['dueDate']))
{
$errorMessage .= "<li>You Forgot To Enter A Due Date !</li>";
}
$lists = $_POST['lists'];
$JN = $_POST['JobName'];
$DES = $_POST['Description'];
$NoS = $_POST['NoStudent'];
$DuDate = $_POST['dueDate'];
if(!empty($errorMessage))
{
echo("<p>There was an error with your form:</p>\n");
echo("<ul>" . $errorMessage . "</ul>\n");
die();
}
}
//////// IF all test passed.. then connect to db
$con = mysqli_connect("localhost","root" ,"" ,"CIE") or die ("cannot connect : ".mysqli_error());
$sql = "INSERT INTO Food (JobName,Description,NoStudent,DueDate)
VALUES
('$JN','$DES','$NoS','$DuDate')";
mysqli_query($con, $sql) or die(mysqli_error($con));
echo "You Successfuly Added a new Reocord ...";
mysqli_close($con);
}
else { echo '
<form action= "AddJob.php" method = "post">
<table width ="100%" cellpadding ="4" border="1" >
<tr>
<th>Select a Catagory</th>
<th>Jobs Name</th>
<th>Description</th>
<th> No Students needed</th>
<th>Due Date</th>
</tr>';
echo "<tr>
<td>".
"<select name = lists >
<option name= nothing value= 0 selected >Choose a Catagory</option>
<option name= nothing value= 1> Advertising </option>
<option name= nothing value= 2> Fiscal </option>
<option name= nothing value= 3> Food </option>
<option name= nothing value= 4> Shopping </option>
<option name= nothing value= 5> Rentals </option>
<option name= nothing value= 6> Setting up </option>
<option name= nothing value= 7> Performances </option>
<option name= nothing value= 8> Registration/Ushering </option>
<option name= nothing value= 9> Master of Ceremonies </option>
<option name= nothing value= 10> Cleaning up </option>
<option name= nothing value= 11> Others </option>
</select>"
." </td>
<td> "."<input type=text name=JobName maxlength=50 placeholder='Enter Job Name '>" ." </td>
<td> "."<input type=text name=Description maxlength=50 placeholder='Enter Description '>" ." </td>
<td> ". "<input type=text name=NoStudent maxlength=50 onkeypress=return isNumber(event) placeholder='ONLY NUMBERS'/>" . "</td>
<td>". "<input type=text name=dueDate maxlength=50 placeholder='YYYY-MM-DD'>" ." </td>
</tr>";
echo '
</table>
<br/>
<div align="center">
<input type="submit" name="add" value="Add" />
<input type="reset" value="Clear" />
</div>
</form>
';
}
}else{echo "must logout to see this page..!!";}
?>
<html>
<head><title> Add.. </title></head>
<br>
<body>
</body>
</html>
答案 0 :(得分：0)
如果您的11个表具有相同的结构（字段名称），您只需替换＆＃34; food＆＃34;表名由变量表示，由＆＃34;开关＆＃34;初始化。 on＆＃34; $ lists＆＃34;可变内容。 一个简单的版本可能是这样的：
$table="";
switch($lists)
{
case 1:
$table ="Advertising";
break;
case 2:
$table ="fiscal";
break;
.........other cases..........
case 0;
default;
echo 'Unsupported category';
break;
}
if ($table != ""){
$sql = "INSERT INTO ". $table . "(JobName,Description,NoStudent,DueDate)
VALUES
('$JN','$DES','$NoS','$DuDate')";
}
答案 1 :(得分：0)
这将变得罗嗦，但我保证最后会有代码。
假设您实际上有11个表，@ dev35000在构建SQL查询时使用PHP变量的解决方案适合您。
但是，由于所有表似乎都包含相同类型的数据，因此您可以通过创建具有CategoryId的表来使自己更轻松。你可以把那个表称为Jobs（或者对你有意义的任何东西。我说就是你正在存储的那些工作）然后你会创建一个表来存储所有不同的类别。
您最终得到的表格是：
`Jobs` Table
-----------------------
JobId | JobName | Description | NoStudent | DueDate | CategoryId
INT, PK, AI | VARCHAR(50) | VARCHAR(50) | VARCHAR(50) | VARCHAR(50) | INT
-----------------------
Filled with data inserted through the form
`Category` Table
-----------------------
CategoryId | Category
INT, PK, AI | VARCHAR(50)
-----------------------
1 | Advertising
2 | Fiscal
3 | Food
4 | Shopping
5 | Rentals
6 | Setting up
7 | Performances
8 | Registration/Ushering
9 | Master of Ceremonies
10 | Cleaning up
11 | Others
JobId和CategoryId将为auto_incrmenting columns
您已经发送了CategoryId，因此您的代码实际上不会在那里发生变化。要显示Category表中Job表记录所属的类别，可以使用SQL中的JOIN来显示作业所属的类别，并使用以下查询：
SELECT j.JobName, j.Description, j.NoStudent, j.DueDate, c.Category
FROM Jobs j
JOIN Category c ON j.CategoryId = c.CategoryId
在我们开始使用代码之前：
isset($_POST['add'])进行了2次检查。删除内部的
因为它是多余的<html>
<head><title> Add.. </title></head>
<body>
<?php
session_start();
if(isset($_SESSION['username'])) {
if(isset($_POST['add'])) {
$errorMessage = "";
if(($_POST['lists'])== 0) { // trying to get error if user don't choose.
$errorMessage .= "<li>You Forgot to Choose !</li>";
}
if(empty($_POST['JobName'])) {
$errorMessage .= "<li>You Forgot To Enter A Job Name !</li>";
}
if(empty($_POST['Description'])) {
$errorMessage .= "<li>You Forgot To Enter A Description !</li>";
}
if(empty($_POST['NoStudent'])) {
$errorMessage .= "<li>You Forgot To Enter A Student Number !</li>";
}
if(empty($_POST['dueDate'])) {
$errorMessage .= "<li>You Forgot To Enter A Due Date !</li>";
}
if(!empty($errorMessage)) {
echo("<p>There was an error with your form:</p>\n");
echo("<ul>" . $errorMessage . "</ul>\n");
die();
}
$JobName = $_POST['JobName'];
$Description = $_POST['Description'];
$NoStudent = $_POST['NoStudent'];
$DueDate = $_POST['dueDate'];
$lists = $_POST['lists'];
//////// IF all test passed.. then connect to db
$con = mysqli_connect("localhost","root" ,"" ,"CIE") or die ("cannot connect : ".mysqli_error());
$sql = "INSERT INTO Jobs (JobName,Description,NoStudent,DueDate,Category)
VALUES
(?, ?, ?, ?, ?, ?)";
// reference: http://php.net/manual/en/mysqli.prepare.php
if ($stmt = mysqli_prepare($con, $sql)) {
// reference: http://php.net/manual/en/mysqli-stmt.bind-param.php
mysqli_stmt_bind_param($stmt, 'ssisi', $JobName, $Description, $NoStudent, $DueDate, $lists);
mysqli_execute($con, $sql) or die(mysqli_error($con));
echo "You Successfuly Added a new Reocord ...";
}
mysqli_close($con);
}
else {
echo "
<form action= 'AddJob.php' method = 'post'>
<table width ='100'% cellpadding ='4' border='1' >
<tr>
<th>Select a Category</th>
<th>Jobs Name</th>
<th>Description</th>
<th>No Students needed</th>
<th>Due Date</th>
</tr>
<tr>
<td>
<select name = lists >
<option name= nothing value= 0 selected >Choose a Category</option>
<option name= nothing value= 1> Advertising </option>
<option name= nothing value= 2> Fiscal </option>
<option name= nothing value= 3> Food </option>
<option name= nothing value= 4> Shopping </option>
<option name= nothing value= 5> Rentals </option>
<option name= nothing value= 6> Setting up </option>
<option name= nothing value= 7> Performances </option>
<option name= nothing value= 8> Registration/Ushering </option>
<option name= nothing value= 9> Master of Ceremonies </option>
<option name= nothing value= 10> Cleaning up </option>
<option name= nothing value= 11> Others </option>
</select>
</td>
<td> <input type=text name=JobName maxlength=50 placeholder='Enter Job Name '></td>
<td> <input type=text name=Description maxlength=50 placeholder='Enter Description '></td>
<td> <input type=text name=NoStudent maxlength=50 onkeypress=return isNumber(event) placeholder='ONLY NUMBERS'/></td>
<td> <input type=text name=dueDate maxlength=50 placeholder='YYYY-MM-DD'></td>
</tr>
</table>
<br/>
<div align='center'>
<input type='submit' name='add' value='Add' />
<input type='reset' value='Clear' />
</div>
</form>";
}
}
else
{
echo "must login to see this page..!!";
}
?>
</body>
</html>
如果我能搞清楚的话，请告诉我。