如何在表格中添加表单,并提交MYSQL数据库 mysqlphp 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 我在你基礎上修改了一下,思路就是這樣,用表記錄的id對應提交。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); // 跳轉回去?> 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='.$q); // 跳轉回去?></body></html>现在运行后出现错误:Parse error: syntax error, unexpected T_VARIABLE in D:\AppServ\www\add.php on line 24请问怎么解决? 24行的 盈利额='"$_POST['yl'.$row['id'].]"' where 改为 盈利额='".$_POST['yl'.$row['id']]."' where 本帖最后由 xuzuning 于 2014-12-10 18:33:58 编辑 感谢各位的帮助,但现在还是有问题:我数据库org中有两个表:1.user:里面有主管工号,姓名,密码,用来作为登陆的账号和密码2.org:里面有id(本来没这个字段,后来加上去的,为了配合傲雪星枫给的代码,自增加字段),架构,盈利额现在用了大家给我的代码后,还是无法更新数据库的盈利额值,并且add.php中最后加上跳转页面的那个代码的话,实际网页跳转的并非是原先主管登陆的welcome.php界面,还请解答。 感谢各位的帮助,但现在还是有问题:我数据库org中有两个表:1.user:里面有主管工号,姓名,密码,用来作为登陆的账号和密码2.org:里面有id(本来没这个字段,后来加上去的,为了配合傲雪星枫给的代码,自增加字段),架构,盈利额现在用了大家给我的代码后,还是无法更新数据库的盈利额值,并且add.php中最后加上跳转页面的那个代码的话,实际网页跳转的并非是原先主管登陆的welcome.php界面,还请解答。 各位前辈,目前问题就出在这段代码上,还请指点如何修改才能根据数据库中id对应的盈利额进行更新:if($_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。 关于文件上传目录创建的问题 curl问题,在线 香港IT界 数组转换,急... mysql如何插入hmtl代码? PHP的GD库问题 PHP实现IP自动判断城市功能 一个关于文本当中的PHP数组问题. 大家看看这个怪问题!!!! 批量向html中插入内容 关于前后台,求帮忙,在线等 循环问题求大神解答 我的QQ是2509577255
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); // 跳轉回去
?>
<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='.$q); // 跳轉回去
?>
</body>
</html>现在运行后出现错误:Parse error: syntax error, unexpected T_VARIABLE in D:\AppServ\www\add.php on line 24
请问怎么解决?
盈利额='"$_POST['yl'.$row['id'].]"' where
改为
盈利额='".$_POST['yl'.$row['id']]."' where
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。
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。
$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());
}
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。