CREATE TABLE IF NOT EXISTS `penyakitt` (
`id_penyakitt` int(11) NOT NULL,
`namapenyakit` varchar(200) NOT NULL,
PRIMARY KEY (`namapenyakit`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE IF NOT EXISTS `cfs` (
`id_cfs` int(11) NOT NULL AUTO_INCREMENT,
`namapenyakit` varchar(200) NOT NULL,
`namagejalaa` varchar(200) NOT NULL,
`mb` varchar(200) NOT NULL,
`md` varchar(200) NOT NULL,
PRIMARY KEY (`id_cfs`),
KEY `namapenyakit` (`namapenyakit`),
KEY `namagejalaa` (`namagejalaa`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=14 ;
CREATE TABLE IF NOT EXISTS `gejalaa` (
`id_gejala` int(11) NOT NULL,
`namagejalaa` varchar(200) NOT NULL,
PRIMARY KEY (`namagejalaa`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Form_input.php
Form_input.php
<form action="/tugasakhir/nilaicf/simpan_nilaicf.php" method="post" >
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="24%">Jenis Penyakit</td>
<td width="5%">:</td>
<td width="71%"><label for="namapenyakit"></label>
<select id="penyakit" name="penyakit">
<option value="">-Pilih penyakit-</option>
<?php
$sql = "SELECT * FROM penyakitt";
$hasil_query=mysql_query($sql);
while($baris = mysql_fetch_object($hasil_query)){
echo "<option value=$baris->id_penyakitt->$baris->namapenyakit</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>Nama gejala</td>
<td>:</td>
<td>
<label for="namagejala"></label>
<select id="gejala" name="gejala">
<option value="">-Pilih gejala-</option>
<?php
$sql = mysql_query("SELECT * FROM gejalaa ORDER BY namagejalaa ASC");
while($data = mysql_fetch_assoc($sql)){
echo "<option value='$data[id_gejala]'>$data[namagejalaa]</option>";
}
?>
</select>
</tr>
<tr>
<td>Nilai MB</td>
<td>:</td>
<td><label for="MB"></label>
<input type="text" name="MB" id="MB" /></td>
</tr>
<tr>
<td>Nilai MD</td>
<td>:</td>
<td><label for="MD"></label>
<input type="MD" name="MD" id="MD" /></td>
</tr>
<tr>
<td height="41"> </td>
<td> </td>
<td><input type="submit" name="tambah" value="Tambah" /></td>
</tr>
</table>
</form>
simpan_nilaicf.php
simpan_nilaicf.php
<?php
include_once "config.php";
$namapenyakit=$_POST['namapenyakit'];
$namagejalaa=$_POST['namagejalaa'];
$mb=$_POST['mb'];
$md=$_POST['md'];
$sql="INSERT INTO cfs ('namapenyakit', 'namagejalaa', 'mb', 'md') VALUES ('', '$namapenyakit', '$namagejalaa', '$mb', '$md')";
$eksekusi_query=mysql_query($sql);
if(!$eksekusi_query){
die("Query kamu salah dikarenakan:".mysql_error());
}
?>
When i run my insert code in mysql phpmyadmin.. its appear error like this :
当我在mysql phpadmin中运行插入代码时。其出现的错误如下:
#1064 - You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the right syntax to use near ''id_cfs', 'namapenyakit', 'namagejalaa', 'mb', 'md') VALUES ('', '$namapenyakit'' at line 1
What should I change , so the data that I input in form_input.php can insert into my database? :(
我应该改变什么,我在form_input中输入的数据。php可以插入我的数据库吗?:(
3 个解决方案
#1
2
In your insert query you are passing 5 values corresponding to the 4 columns mentioned by you and for columns you are using wrong quotes.
在您的插入查询中,您将传递与您所提到的4列相对应的5个值,以及您使用错误引用的列。
Try the following:
试试以下:
$sql="INSERT INTO cfs (`namapenyakit`, `namagejalaa`, `mb`, `md`) VALUES ('$namapenyakit', '$namagejalaa', '$mb', '$md')";
#2
0
You insert one empty value:
插入一个空值:
$sql="INSERT INTO cfs (namapenyakit, namagejalaa, mb, md) VALUES ('$namapenyakit', '$namagejalaa', '$mb', '$md')";
This is not safe way. Use prepared statements.
这不是安全的方法。使用预处理语句。
#3
0
fyi:
仅供参考:
when i was input manually the data in mysql with this :
当我手工输入mysql数据时:
INSERT INTO cfs
( id_cfs
, namapenyakit
, namagejalaa
, mb
, md
) VALUES ( 3, 'asma', 'muntah disertai riak', '5', '4' )
插入cfs (id_cfs, namapenyakit, namagejalaa, mb, md)值(3,'asma', 'muntah disertai riak', '5', '4')
it can added.. but i still dont know how to make the data that I input in form_input.php can insert into my database...
它可以添加. .但是我仍然不知道如何将输入的数据输入form_input。php可以插入到我的数据库中…
#1
2
In your insert query you are passing 5 values corresponding to the 4 columns mentioned by you and for columns you are using wrong quotes.
在您的插入查询中,您将传递与您所提到的4列相对应的5个值,以及您使用错误引用的列。
Try the following:
试试以下:
$sql="INSERT INTO cfs (`namapenyakit`, `namagejalaa`, `mb`, `md`) VALUES ('$namapenyakit', '$namagejalaa', '$mb', '$md')";
#2
0
You insert one empty value:
插入一个空值:
$sql="INSERT INTO cfs (namapenyakit, namagejalaa, mb, md) VALUES ('$namapenyakit', '$namagejalaa', '$mb', '$md')";
This is not safe way. Use prepared statements.
这不是安全的方法。使用预处理语句。
#3
0
fyi:
仅供参考:
when i was input manually the data in mysql with this :
当我手工输入mysql数据时:
INSERT INTO cfs
( id_cfs
, namapenyakit
, namagejalaa
, mb
, md
) VALUES ( 3, 'asma', 'muntah disertai riak', '5', '4' )
插入cfs (id_cfs, namapenyakit, namagejalaa, mb, md)值(3,'asma', 'muntah disertai riak', '5', '4')
it can added.. but i still dont know how to make the data that I input in form_input.php can insert into my database...
它可以添加. .但是我仍然不知道如何将输入的数据输入form_input。php可以插入到我的数据库中…