如何在表格中添加表单，并提交MYSQL数据库
本人是个新手，现在想做个指标提交系统，MYSQL数据库中已经有销售整体架构，并已经在PHP中按照登陆账号自动生成对应的销售下属人员，但是架构后面跟着销售的指标，想做个输入框，并提交到数据库中，请问怎么做：
架构 盈利额
-L4:上海结婚 黄主管 输入框
-L3:上海结婚2组 沈主管 输入框
-L2:上海结婚2-1组 马主管 输入框
L1: 孙销售 输入框
L1: 杜销售 输入框
L1: 李销售 输入框
<?php $q=$_POST["employeenumber"];$con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("org", $con);echo "<table border='1' cellpadding='10'><tr><th>架构</th><th>盈利额</th></tr>";$sql="SELECT 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'";$result = mysql_query($sql);while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['架构'] . "</td>"; echo "</tr>"; }echo "</table>";mysql_close($con);?><br>
回复讨论(解决方案)
本人是个新手，现在想做个指标提交系统，，请大家做好了，发给我：
我在你基?上修改了一下，思路就是??，用表??的id??提交。
index.php
<?php $q=$_POST["employeenumber"]; $con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("org", $con);echo '<form name="form1" method="post" action="add.php">';echo "<table border='1' cellpadding='10'><tr><th>架构</th><th>盈利额</th></tr>"; $sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['架构'] . "</td>"; echo '<td><input type="text" name="yl'.$row['id'].'"></td>'; echo "</tr>"; }echo "</table>";echo '<input type="hidden" name="employeenumber" value="'.$q.'">';echo '</form>'; mysql_close($con);?>add.php
<?php$con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("org", $con);$employeenumber = $_POST["employeenumber"];$sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)){ if($_POST['yl'.$row['id']]){ $sqlstr = "update `org` set 盈利额='".$_POST['yl'.$row['id']]."' where id='".$row['id']."'"; // 更新入db mysql_query($sqlstr) or die(mysql_error()); }}mysql_close($con);header('location:index.php?q='.$employeenumber); // 跳?回去?>多谢回复，明天我试试看。
可能我没解释清楚，我把整个系统代码结合前辈的代码重新写了下：
登陆界面（未完整）：
<html><title>指标收集系统</title><body><center><form action="welcome.php" method="post">工号: <input type="text" name="employeenumber" placeholder="请输入7位工号"><br>密码: <input type="password" name="password"><br><input type="submit" value="登录"></form></center></body></html>
welcome.php
<html><body>欢迎<?php $q=$_POST["employeenumber"];$con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("org", $con);$sql="SELECT 姓名 FROM user WHERE 员工号 = '".$q."'";$result = mysql_query($sql);$row = mysql_fetch_array($result);echo $row['姓名'];echo '<form name="form1" method="post" action="add.php">';echo "<table border='1' cellpadding='10'><tr><th>架构</th><th>盈利额</th></tr>";$sql="SELECT id,架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'";$result = mysql_query($sql);while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['架构'] . "</td>"; echo '<td><input type="number" name="yl'.$row['id'].'"></td>'; echo "</tr>"; }echo "</table>";//echo '<input type="hidden" name="employeenumber" value="'.$q.'">';echo '<input type="submit" value="提交指标">';mysql_close($con);?><br></body></html>add.PHP
<html><head></head><body><?php$con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("org", $con); //$employeenumber = $_POST["employeenumber"]; $sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)){ if($_POST['yl'.$row['id']]){ $sqlstr = "update `org` set 盈利额='"$_POST['yl'.$row['id'].]"' where id='".$row['id']."'"; // 更新入db mysql_query($sqlstr) or die(mysql_error()); }} mysql_close($con); //header('location:welcome.php?q='.$q); // 跳?回去?></body></html>现在运行后出现错误：Parse error: syntax error, unexpected T_VARIABLE in D:\AppServ\www\add.php on line 24
请问怎么解决？
24行的
盈利额='"$_POST['yl'.$row['id'].]"' where
改为
盈利额='".$_POST['yl'.$row['id']]."' where
$sqlstr = "update `org` set 盈利额='"$_POST['yl'.$row['id'].]"' where id='".$row['id']."'";
应为
$sqlstr = "update `org` set 盈利额='" . $_POST['yl'.$row['id']] . "' where id='".$row['id']."'";
感谢各位的帮助，但现在还是有问题：
我数据库org中有两个表：
1.user：里面有主管工号，姓名，密码，用来作为登陆的账号和密码
2.org：里面有id（本来没这个字段，后来加上去的，为了配合傲雪星枫给的代码，自增加字段），架构，盈利额
现在用了大家给我的代码后，还是无法更新数据库的盈利额值，并且add.php中最后加上跳转页面的那个代码的话，实际网页跳转的并非是原先主管登陆的welcome.php界面，还请解答。
感谢各位的帮助，但现在还是有问题：
我数据库org中有两个表：
1.user：里面有主管工号，姓名，密码，用来作为登陆的账号和密码
2.org：里面有id（本来没这个字段，后来加上去的，为了配合傲雪星枫给的代码，自增加字段），架构，盈利额
现在用了大家给我的代码后，还是无法更新数据库的盈利额值，并且add.php中最后加上跳转页面的那个代码的话，实际网页跳转的并非是原先主管登陆的welcome.php界面，还请解答。
各位前辈，目前问题就出在这段代码上，还请指点如何修改才能根据数据库中id对应的盈利额进行更新：
if($_POST[.$row['id'].]){ $sqlstr = "update `org` set 盈利额='" . $_POST[.$row['id'].] . "' where id='".$row['id']."'";// 更新入db mysql_query($sqlstr) or die(mysql_error()); }去看下jquery easy ui的datagrid很简单，加载数据还可以让它自动增加输入框，只需要一个属性
费了老半天劲，总算自己搞定了，贴上代码：
welcome.php
<html><body> 欢迎<?php $q=$_POST["employeenumber"]; $con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("org", $con); $sql="SELECT 姓名 FROM user WHERE 员工号 = '".$q."'"; $result = mysql_query($sql); $row = mysql_fetch_array($result); echo $row['姓名']; echo '<form name="form1" method="post" action="add.php">'; echo "<table border='1' cellpadding='10'><tr><th>架构</th><th>盈利额</th></tr>"; $sql="SELECT id,架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['架构'] . "</td>"; echo '<td><input type="number" name= '.$row['id'].'></td>'; echo "</tr>"; } echo "</table>";echo '<input type="hidden" name="employeenumber" value="'.$q.'">';echo '<input type="submit" value="提交指标">'; mysql_close($con);?><br> </body></html>add.php
<html><head></head><body> <?php$con = mysql_connect('localhost', 'root', '');if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("org", $con);$employeenumber = $_POST["employeenumber"]; $sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$employeenumber."'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)){ if($_POST[$row['id']]){ $sqlstr = "update `org` set 盈利额='".$_POST[$row['id']]."' where id='".$row['id']."' and 主管工号 = '".$employeenumber."'"; // 更新入db mysql_query($sqlstr) or die(mysql_error()); }} mysql_close($con);echo "指标提交成功";?> </body></html>还是很感谢各位的帮助。
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