我尝试将数据学生名称'Dhiya'和'Siti'和$ checkbox插入数据库。我首先从数据库中的student_course表中回忆起数据。我检查了Presence的名字。这是代码:
<html>
<div class="container">
<div class="content">
<?php
include ('system2.php');
include('connect_database.php');
$course = $_POST['course'];
$batch = $_POST['batch'];
$group = $_POST['group'];
$student_name = $_POST['Student_Name'];
$checkbox = $_POST['checkbox'];
$query_view1 = mysql_query("SELECT distinct Student_Name FROM student_course where batch='12/01'");
$query_view2 = mysql_query("SELECT distinct Student_Name FROM student_course where batch='12/02'");
echo "<form action ='system23.php' method = 'post'>";
echo "<table border = '1' align='center'>";
echo "<tr>
<th><font color=#555555>Name</font> </th>
<th><font color=green>Presence</font> </th>
<th><font color=red>Absence</font> </th>
<th><font color=#FFCC00>Permission</font> </th>
<th><font color=purple>Late</font> </th>
</tr>";
if ($batch=='12/01')
{
while($row = mysql_fetch_array($query_view1) )
{
echo "<tr>";?>
<td><input type="text" name="stuname" value="<?php echo $row['Student_Name']?>"/></td>
<?php
echo "<td><input type='checkbox' name='checkbox' value='P'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='A'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='PR'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='L'/></td>";
echo "</tr>";
}
}
else if ($batch=='12/02')
{
while($row = mysql_fetch_array($query_view2) )
{
echo "<tr>";?>
<td><input type="text" name="stuname" value="<?php echo $row['Student_Name']?>"/></td>
<?php
echo "<td><input type='checkbox' name='checkbox' value='P'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='A'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='PR'/></td>";
echo "<td><input type='checkbox' name='checkbox' value='L'/></td>";
echo "</tr>";
}
}
echo "<tr>";
echo "<td colspan='6' align='center'><input name='submit' type='submit' value='GO' /></td>";
echo "</tr>";
echo "</table>";
echo "</form>";
?>
</div>
</div>
</html>
然后我尝试将数据插入数据库。这是代码:
<html>
<div class="container">
<?php
include ('system22.php');
include('connect_database.php');
$checkbox = $_POST['checkbox'];
$query_view1 = mysql_query("SELECT distinct Student_Name FROM student_course where batch='12/02'");
$query_view2 = mysql_query("SELECT distinct Student_Name FROM student_course where batch='12/02'");
$student_name = $_POST['stuname'];
$batch= $_POST['batch'];
mysql_query ("INSERT INTO result2 VALUES ('$student_name', '$checkbox')");
echo "Success";
?>
</div>
</html>
但是当我检查数据库时,它只放了'Siti'。我不知道还有其他什么方法可以使用
由于我无法发布图片,我会尽量让它尽可能清晰
student_course表
<html>
<table>
<tr>
<th>Name</th>
<th>1</th>
</tr>
<tr>
<td>Siti</td>
<td>P</td>
</tr>
</table>
</html>
对于表result2结构。
字段:'名称'和'1' 类型:两个varchar(99)。 没有主要和前卫的钥匙。
答案 0 :(得分:1)
如果您只将所有内容插入一列,则需要在整个字符串周围加上引号,而不是两个变量。
mysql_query ("INSERT INTO result2 VALUES ('$student_name, $checkbox')");
应该有效
或者您可以简单地将两个变量添加到新的变量中,如:
$insert = $student_name.", ".$checkbox;
mysql_query ("INSERT INTO result2 VALUES ('$insert')");
但是他们最好的选择是切换到PDO
例如并准备语句以避免SQL
一劳永逸地注入;)
答案 1 :(得分:0)
例如,对天气使用if语句或不选中该框。
if (!isset($_POST['checkboxname'])) {
// checkbox was not checked
} else {
// checkbox was checked
}