2013-09-24 39 views
-1

以下是用戶在其中輸入數據的表格。如何在兩張不同的表格中使用單一格式插入表格數據

<form action="addemployee.php" method="POST" id="addEmployeeToDataBase"> 
    <input type="text" name="username" placeholder="UserName..." class="txtInput" required="" /> 
    <input type="text" name="fullname" placeholder="Full Name..." class="txtInput" required="" /> 
    <br /> 
    <input type="email" name="email" placeholder="Email Address..." class="txtInput" required="" /> 
    <input type="password" name="password" placeholder="Password..." class="txtInput" required="" /> 
    <br /> 
    <select name="dept" id="txtInput"> 
     <option value="-1">Select Department</option> 
     <option value="Back-Office">Back-Office</option> 
     <option value="HR">HR</option> 
     <option value="Marketing">Marketing</option> 
     <option value="Sales">Sales</option> 
     <option value="Technical">Technical</option> 
    </select> 
    <br /> 
    <input type="text" name="address" placeholder="Address..." class="txtInput" required=""/> 
    <input type="text" name="city" placeholder="City..." class="txtInput" required="" /> 
    <br /> 
    <input type="text" name="pincode" placeholder="Pin/Postal/Zip Code..." class="txtInput" required="" /> 
    <input type="text" name="country" placeholder="Country..." class="txtInput" required="" /> 
    <br /> 
    <p style="font-size: 17px;" />Date of Joining:<input type="date" name="joiningdate" placeholder="Joining Date" class="txtInput" required="" /> 
    <br /> 
    <input type="submit" name="submit" id="submit" value="Add Employee"/> 

現在我想使用SQL語句在輸入的數據將同時發送到不同的表。

Table employee (emp_id, username, email, password, dept, address, city, pincode, country and joiningdate). 

Table dept (dept_id, dept_name, dept_head, and emp_id). 

該表格基本上插入到僱員表中。

現在,「問題」是我想分別將單個表單的數據插入到employee表和dept表中,其中表單中dept的輸入數據也應該插入到dept表的dept_name列中,並且還應該更新輸入數據的dept表的emp_id。

請指導我完成此操作。

+0

是您的EMP_ID自動遞增? – Salim

+0

是的,它是auto_incremented –

+0

好吧然後相關的答案已經給他人 – Salim

回答

1

首先將數據插入到employee表中,然後使用mysql函數last_insert_id可以獲取emp_id值並將其插入到department表中。

假設,emp_id是自動遞增的值。

2
$query1= "INSERT INTO employee (username, email,...) 
      VALUES ('".$_POST["username"]."', ...)"; 
     if($result1 = mysql_query($query1)) 
     { 
      $emp_id = mysql_insert_id(); 

      $query2= "INSERT INTO dept (emp_id, dept_name, ...) 
      VALUES ('".$emp_id."', '".$_POST["dept_name"]."',...)"; 

      if($result2 = mysql_query($query2)) 
      { 
       //success msg 
      } 
     } 
+0

謝謝老兄!你真的幫了我很多.. –

0

這可能幫助(我使用Ubuntu 12.04和LAMP它爲我工作)

$sql1 = "INSERT INTO users_account_details (uname,email,pass1,pass2) 
VALUES('$_POST[uname]','$_POST[email]','$_POST[pass1]',DES_ENCRYPT('$_POST[pass2]'))"; 

$sql2 = "INSERT INTO users_personal_details (name,gender,dob,address,mobile) 
VALUES('$_POST[name]','$_POST[gender]','$dob','$_POST[address]','$_POST[mobile]')"; 

if (!mysqli_query($con,$sql1)) 
    { 
    die('Error: ' . mysqli_error($con)); 
    } 
if (!mysqli_query($con,$sql2)) 
    { 
    die('Error: ' . mysqli_error($con)); 
} 
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