php - 无法获取表单来更新 mysql 记录

Question

我有一个由三部分组成的表单，其工作原理如下，第一个表单称为 create_ticket.php，它基本上是一个包含客户联系信息和他们想要完成的操作的表单。填写完票证后，它会提交到 MySQL，然后当数据库搜索完成后，您可以通过单击搜索结果中与该票证关联的选择链接来选择所需的票证，浏览器将打开一个新窗口，其中包含表单的第二部分，但它将传递与浏览器中的 MySQL 记录关联的 id，因此您可以在 url 末尾看到 ?id=10 现在，表单的第二部分是技术人员将用来填写的内容他们这样做了，当他们点击页面底部的“获取签名”时，会弹出表单的第三部分，允许客户在技术人员的 ipad 或 Toughbook 上签名，当他们点击“接受”时，它应该更新与关联的 MySQL 中的记录该 id 在 url 中传递。我的问题是它不起作用。这样您就可以在 http://jemtechnv.com/test/ticket_results.php 查看表格当您单击选择时，您会看到需要填写的技术人员表格。 这是我处理表单第二部分的代码:

<?php
// database connection //

include 'db/db_connect.php';

//This gets all the other information from the form

// start of form inputs //

$work_performed=$_POST['work_performed'];
$item_qty1=$_POST['item_qty1'];
$item_qty2=($_POST['item_qty2']);
$item_qty3=$_POST['item_qty1'];
$item_qty4=($_POST['item_qty2']);
$item_qty5=$_POST['item_qty1'];
$manuf_1=$_POST['manuf_1'];
$manuf_2=$_POST['manuf_2'];
$manuf_3=$_POST['manuf_3'];
$manuf_4=$_POST['manuf_4'];
$manuf_5=$_POST['manuf_5'];
$part_number1=$_POST['part_number1'];
$part_number2=$_POST['part_number2'];
$part_number3=$_POST['part_number3'];
$part_number4=$_POST['part_number4'];
$part_number5=$_POST['part_number5'];
$part_description1=$_POST['part_description1'];
$part_description2=$_POST['part_description2'];
$part_description3=$_POST['part_description3'];
$part_description4=$_POST['part_description4'];
$part_description5=$_POST['part_description5'];
$part1_price=$_POST['part_price1'];
$part2_price=$_POST['part_price2'];
$part3_price=$_POST['part_price3'];
$part4_price=$_POST['part_price4'];
$part5_price=$_POST['part_price5'];
$price_extension1=$_POST['price_extension1'];
$price_extension2=$_POST['price_extension2'];
$price_extension3=$_POST['price_extension3'];
$price_extension4=$_POST['price_extension4'];
$price_extension5=$_POST['price_extension5'];
$material_total=$_POST['material_total'];
$sales_tax=$_POST['sales_tax'];
$shipping_cost=$_POST['shipping_cost'];
$work_date1=$_POST['work_date1'];
$work_date2=$_POST['work_date2'];
$work_date3=$_POST['work_date3'];
$work_date4=$_POST['work_date4'];
$work_date5=$_POST['work_date5'];
$tech_name1=$_POST['tech_name1'];
$tech_name2=$_POST['tech_name2'];
$tech_name3=$_POST['tech_name3'];
$tech_name4=$_POST['tech_name4'];
$tech_name5=$_POST['tech_name5'];
$cost_code1=$_POST['cost_code1'];
$cost_code2=$_POST['cost_code2'];
$cost_code3=$_POST['cost_code3'];
$cost_code4=$_POST['cost_code4'];
$cost_code5=$_POST['cost_code5'];
$pay_rate1=$_POST['pay_rate1'];
$pay_rate2=$_POST['pay_rate2'];
$pay_rate3=$_POST['pay_rate3'];
$pay_rate4=$_POST['pay_rate4'];
$pay_rate5=$_POST['pay_rate5'];
$total_hours1=$_POST['total_hours1'];
$total_hours2=$_POST['total_hours2'];
$total_hours3=$_POST['total_hours3'];
$total_hours4=$_POST['total_hours4'];
$total_hours5=$_POST['total_hours5'];
$hours_subtotal1=$_POST['hours_subtotal1'];
$hours_subtotal2=$_POST['hours_subtotal2'];
$hours_subtotal3=$_POST['hours_subtotal3'];
$hours_subtotal4=$_POST['hours_subtotal4'];
$hours_subtotal5=$_POST['hours_subtotal5'];
$total_hours=$_POST['total_hours'];
$material_total=$_POST['material_total'];
$labor_cost=$_POST['labor_cost'];
$grand_total=$_POST['grand_total'];

//Writes the information to the database
 mysql_query("UPDATE INTO tickets WHERE id=$id (work_performed, item_qty1, item_qty2, item_qty3, item_qty4, item_qty5,manuf_1, manuf_2, manuf_3, manuf_4, manuf_5, part_number1, part_number2, part_number3, part_number4, part_number5, part_description1, part_description2, part_description3, part_description_4, part_description_5, part1_price, part2_price, part3_price, part4_price, part5_price, price_extension1, price_extension2, price_extension3, price_extension4, price_extension5, material_total, sales_tax, shipping_cost, work_date1, work_date2, work_date3, work_date4, work_date5, tech_name1, tech_name2, tech_name3, tech_name4, tech_name5, cost_code1, cost_code2, cost_code3, cost_code4, cost_code5, pay_rate1, pay_rate2, pay_rate3, pay_rate4, pay_rate5, total_hours1, total_hours2, total_hours3, total_hours4, total_hours5, hours_subtotal1, hours_subtotal2, hours_subtotal3, hours_subtotal4, hours_subtotal5, total_hours, material_total, labor_cost, grand_total,)
VALUES ('$work_performed','$item_qty1','$item_qty2','$item_qty3','$item_qty4','$item_qty5','$mauf_1','$manuf_2','$manuf_3','$manuf_4','$manuf_5','$part_number1','$part_number2','$part_number3','$part_number4','$part_number5','$part_description1','$part_description2','$part_description3','$part_description_4',
'$part_description5','$part1_price','$part2_price','$part3_price','$part4_price','$part5_price','$price_extension1','$price_extension2','$price_extension3','$price_extension4','$price_extension5','$material_total','$sales_tax','$shipping_cost','$work_date1','$work_date2','$work_date3','$work_date4','$work_date5','$tech_name1','$tech_name2','$tech_name3','$tech_name4','$tech_name5','$cost_code1','$cost_code2','$cost_code3','$cost_code4','$cost_code5','$pay_rate1','$pay_rate2','$pay_rate3','$pay_rate4','$pay_rate5',
'$total_hours1','$total_hours2','$total_hours3','$total_hours4','$total_hours5','$hours_subtotal1','$hours_subtotal2','$hours_subtotal3','$hours_subtotal4','$hours_subtotal5','$total_hours','$material_total','$labor_cost','$grand_total',)") ; 



mysql_affected_rows();

    ?>
<html>
<body>
<center>
<br><br><br>
<form name="results" method="post" action="ticket_results.php" enctype="multipart/form-data" id="ticketresult">
<input type="submit" class="submit" id="ticketresult" style="width: 165px" value="Do Something">
</form>
</center>
</body>
</html>

我也知道我需要采取 MySQL 注入(inject)预防措施，但老实说，目前这不是优先事项，因为我只想先让表单正常工作，然后我会返回并清理代码。感谢您的帮助!我已经禁用了表单的签名部分，因此单击第二个表单上的“获取签名”将处理该表单并对该记录进行更新，这是应该的，但事实并非如此!

更新:我已将查询语法更改为以下建议:

mysql_query("UPDATE tickets SET (work_performed, item_qty1, item_qty2, item_qty3, item_qty4, item_qty5,manuf_1, manuf_2, manuf_3, manuf_4, manuf_5, part_number1, part_number2, part_number3, part_number4, part_number5, part_description1, part_description2, part_description3, part_description_4, part_description_5, part1_price, part2_price, part3_price, part4_price, part5_price, price_extension1, price_extension2, price_extension3, price_extension4, price_extension5, material_total, sales_tax, shipping_cost, work_date1, work_date2, work_date3, work_date4, work_date5, tech_name1, tech_name2, tech_name3, tech_name4, tech_name5, cost_code1, cost_code2, cost_code3, cost_code4, cost_code5, pay_rate1, pay_rate2, pay_rate3, pay_rate4, pay_rate5, total_hours1, total_hours2, total_hours3, total_hours4, total_hours5, hours_subtotal1, hours_subtotal2, hours_subtotal3, hours_subtotal4, hours_subtotal5, total_hours, material_total, labor_cost, grand_total,)WHERE id=$id
VALUES ('$work_performed','$item_qty1','$item_qty2','$item_qty3','$item_qty4','$item_qty5','$mauf_1','$manuf_2','$manuf_3','$manuf_4','$manuf_5','$part_number1','$part_number2','$part_number3','$part_number4','$part_number5','$part_description1','$part_description2','$part_description3','$part_description_4',
'$part_description5','$part1_price','$part2_price','$part3_price','$part4_price','$part5_price','$price_extension1','$price_extension2','$price_extension3','$price_extension4','$price_extension5','$material_total','$sales_tax','$shipping_cost','$work_date1','$work_date2','$work_date3','$work_date4','$work_date5','$tech_name1','$tech_name2','$tech_name3','$tech_name4','$tech_name5','$cost_code1','$cost_code2','$cost_code3','$cost_code4','$cost_code5','$pay_rate1','$pay_rate2','$pay_rate3','$pay_rate4','$pay_rate5',
'$total_hours1','$total_hours2','$total_hours3','$total_hours4','$total_hours5','$hours_subtotal1','$hours_subtotal2','$hours_subtotal3','$hours_subtotal4','$hours_subtotal5','$total_hours','$material_total','$labor_cost','$grand_total',)"); 

但是现在当我提交表单更新数据库时出现以下错误，

您的 SQL 语法有错误；检查与您的 MySQL 服务器版本对应的手册，了解在第 1 行 '(work_performed, item_qty1, item_qty2, item_qty3, item_qty4, item_qty5,manuf_1, ' 附近使用的正确语法

Answer 1

您的查询有误。查询应该是

UPDATE tickets  SET work_performed = something, item_qty1 = something .... WHERE id=$id