问题描述
我正在尝试使用一个 html 表单来更新 mysql 数据.现在,我有这个代码(这也是一个表单操作),我也试图将它用作我的更新表单.因为我将需要此表单将显示的数据,以便用户更轻松地仅更新他们希望更新的内容.
I'm trying to have an html form which updates mysql data. Now , I have this code(which is also a form action) and I'm trying to also use this as a form for my update. Because I will need the data that this form would show, so that it will be easier for the users to update only what they wish to update.
这是将尝试搜索数据的表单:
this is the form that will try to search the data :
<form name="form1" method="post" action="new.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="16" style="background:#9ACD32; color:white; border:white 1px solid;
text-align: center"><strong><font size="3">ADMISSION INFORMATION SHEET</strong></td>
</tr>
<tr>
这是 new.php( 将根据输入的名字显示相应的数据.并且还将尝试作为更新过程的表单.
This is new.php( will display the corresponding data based on the firstname inputted. And will also try to serve as a form for the update process.
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("Hospital", $con);
$result = mysql_query("SELECT * FROM t2 WHERE FIRSTNAME='{$_POST["fname"]}'");
?>
<table width="900" border="0" align="left" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="16" style="background:#9ACD32; color:white; border:white 1px solid; text-align: center"><strong><font size="3">ADMISSION INFORMATION SHEET</strong></td>
</tr>
<tr>
<?php while ( $row = mysql_fetch_array($result) ) { ?>
<form name="form1" method="post" action="update.php">
<td width="54"><font size="3">Hospital #</td>
<td width="3">:</td>
<td width="168"><input name="hnum" type="text" value="<?php echo $row["HOSPNUM"]; ?>">
</td>
这是我的update.php,
This is my update.php,
mysql_select_db("Hospital", $con);
mysql_query("UPDATE t2 SET HOSPNUM='$_POST[hnum]' ROOMNUM='$_POST[rnum]',
LASTNAME='$_POST[lname]', FIRSTNAME='$_POST[fname]', MIDNAME='$_POST[mname]',
CSTAT='$_POST[cs]' AGE='$_POST[age]', BDAY='$_POST[bday]', ADDRESS='$_POST[ad]',
STAT='$_POST[stats1]', STAT2'$_POST[stats2]', STAT3'$_POST[stats3]',
STAT4'$_POST[stats4]', STAT5'$_POST[stats5]', STAT6'$_POST[stats6]',
STAT7'$_POST[stats7]', STAT8'$_POST[stats8]', NURSE='$_POST[nurse]', TELNUM
='$_POST[telnum]'
WHERE FNAME ='$_POST[fname]'");
mysql_close($con);
?>
-请帮忙,我不知道为什么它不更新数据.
-Please help, I don't have any idea why it isnt updating the data.
推荐答案
打字错误,HOSPNUM 和 ROOMNUM 之间缺少,":SET HOSPNUM='$_POST[hnum]', ROOMNUM=
Typo, there is a missing "," between HOSPNUM and ROOMNUM:SET HOSPNUM='$_POST[hnum]', ROOMNUM=
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