從動態文本字段將數據插入mysql，然后將數據插入兩個表中。
[英]insert data into mysql from the dynamic text field and have ti insert the data into two tables.
問題描述
我有2個數據庫，一個是backup_list和磁帶。 backup_id是backup_list的PK，是磁帶的FK。 如何將來自一個查詢的數據插入到兩個表中。 有用於文本代碼，備注和驗證的動態文本字段。 請幫我。
<script type="text/javascript"> var i = 0; $(document).ready(function() { $('.addNewTape').on( "click", function() { $(this).parent().siblings(".Tape_Code").append('<input type="text" name="inputTapeCode[1]['+ i +']" value="'+ i +'" /><br>'); $(this).parent().siblings(".Operator_Verify_Status").append('<SELECT name="Operator_Verify_Status"><OPTION value="verify">Verify</OPTION><OPTION value="NotVerify">Not Verify</OPTION><OPTION value="Failed">Failed</OPTION></SELECT><br>'); $(this).parent().siblings(".Operator_Remark").append('<input type="text" name="inputTapeRemark[1]['+ i +']" value="" /><br>'); i++; }); }); </script> <script> n = new Date(); y = n.getFullYear(); m = n.getMonth() + 1; d = n.getDate(); document.getElementById("date").innerHTML = d + "." + m + "." + y; function addRow(tableID) { var table = document.getElementById(tableID); var rowCount = table.rows.length; var row = table.insertRow(rowCount); var colCount = table.rows[0].cells.length; for(var i=0; i<colCount; i++) { var newcell = row.insertCell(i); newcell.innerHTML = table.rows[0].cells[i].innerHTML; //alert(newcell.childNodes); switch(newcell.childNodes[0].type) { case "text": newcell.childNodes[0].value = ""; break; case "checkbox": newcell.childNodes[0].checked = false; break; case "select-one": newcell.childNodes[0].selectedIndex = 0; break; case "select-one": newcell.childNodes[0].selectedIndex = 0; break; case "select-one": newcell.childNodes[0].selectedIndex = 0; break; case "select-one": newcell.childNodes[0].selectedIndex = 0; break; } } } function deleteRow(tableID) { try { var table = document.getElementById(tableID); var rowCount = table.rows.length; for(var i=0; i<rowCount; i++) { var row = table.rows[i]; var chkbox = row.cells[0].childNodes[0]; if(null != chkbox && true == chkbox.checked) { if(rowCount <= 1) { alert("Cannot delete all the rows."); break; } table.deleteRow(i); rowCount--; i--; } } }catch(e) { alert(e); } } $(document).on('click', '.btn_delete', function(){ var Backup_Name=$(this).data("id5"); if(confirm("Are you sure you want to delete this?")) { $.ajax({ url:"delete.php", method:"POST", data:{Backup_Name:Backup_Name}, dataType:"text", success:function(data){ alert(data); fetch_data(); } }); } }); </script>
<html> <body> <div class="right_col" role="main"> <h1>Daily Backup</h1> <h2 id="date"></h2> <script src="https://code.jquery.com/jquery-3.1.1.min.js"></script> <?php $Date=date('d/m/y'); include_once 'dpconnect.php'; $que=mysqli_query($MySQLiconn,"SELECT Backup_Name FROM admin_backup_list "); ?> <!--set validation error flag as false--> <form name="form2" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <table> <tr> <td>Select</td> <td width="103">Date</td> <td width="94">Backup_Name</td> <td width="94">No Of Tapes</td> <td width="53">Tape Code</td> <td width="71">Operator Approval</td> <td width="144">Operator Remark</td> <td width="144">Submit</td> </tr> <?php if ($que->num_rows > 0) { while ($row = mysqli_fetch_array($que)) { ?> <tr> <td></td> <td><?php echo $Date;?></td> <td><?php echo $row['Backup_Name']; ?></td> <td><input type ="text" name="No_Of_Backup"/><a href="#" class="addNewTape">Add</a></td> <td class="Tape_Code"></td> <td class="Operator_Verify_Status"></td> <td class="Operator_Remark"></td> <!--<td><input type= "button" value="Submit" name="Submit"/></td>--> </tr> <?php } } ?> </table> </form> <form name="form2" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <TABLE id="dataTable"> <TR> <TD><INPUT type="checkbox" name="chk"/></TD> <TD><?php echo $Date;?></TD> <TD><INPUT type="text" name="Backup_Name"/></TD> <TD><INPUT type="text" name="No_Of_Backup"/></TD> <TD><INPUT type="text" name="Tape_Code"/></TD> <TD> <SELECT name="Operator_Verify_Status"> <OPTION value="verify">Verify</OPTION> <OPTION value="NotVerify">Not Verify</OPTION> <OPTION value="Failed">Failed</OPTION> </SELECT> </TD> <TD><INPUT type="text" name="Operator_Remark"/></TD> <td><input type= "submit" value="submit" name="submit"/></td> </TR> </table> <INPUT type="button" value="Add Row" onclick="addRow('dataTable')" /> <INPUT type="button" value="Delete Row" onclick="deleteRow('dataTable')" /> </form> <?php include_once 'dpconnect.php'; //set validation error flag as false $error = false; //check if form is submitted if (isset($_POST['submit'])) { $Date=date('d/m/y'); $Backup_Name = mysqli_real_escape_string($MySQLiconn , $_POST['Backup_Name']); $No_Of_Backup = mysqli_real_escape_string($MySQLiconn , $_POST['No_Of_Backup']); $Tape_Code = mysqli_real_escape_string($MySQLiconn , $_POST['Tape_Code']); $Operator_Verify_Status = mysqli_real_escape_string($MySQLiconn , $_POST['Operator_Verify_Status']); $Operator_Remark = mysqli_real_escape_string($MySQLiconn , $_POST['Operator_Remark']); if (!$error) { if(mysqli_query($MySQLiconn , "INSERT INTO backup_list(Backup_Name, No_Of_Backup) VALUES('" . $Backup_Name . "', '" . $No_Of_Backup . "')")) { $successmsg = "Successfully Registered!"; } else { $errormsg = "Error in registering...Please try again later!"; } } } ?> </div> </body> </html>
1 個解決方案
解決方案1
0 已采納 2016-11-17 05:45:48
使用LAST_INSERT_ID()獲取插入到backup_list時分配的backup_id 。
if(mysqli_query($MySQLiconn , "INSERT INTO backup_list(Backup_Name, No_Of_Backup) VALUES('" . $Backup_Name . "', '" . $No_Of_Backup . "')")) {
mysqli_query($MySQLiconn, "INSERT INTO tapes(backup_id, Tape_Code, Operator_Verify_Status, Operator_Remark)
VALUES (LAST_INSERT_ID(), '$Tape_Code', '$Operator_Verify_Status', '$Operator_Remark')";
$successmsg = "Successfully registered";
}
順便說一句，您應該了解准備好的查詢，而不是在查詢中使用mysqli_real_escape_string和變量替換。 請參閱如何防止在PHP中進行SQL注入？
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