[英]how to insert the checked value of radio button which is array from another table in php
我创建了一个测试表单,试图将来自性别表(包含:(1)男性(2)女性)的单选按钮值发送到另一个名为ajebaje
MySQL表中。 我目前遇到问题。 下面的代码只是一个测试,我希望单选按钮提交该值,但不是。
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title> ajebaje </title>
</head>
<body>
<form name="form1" action= "<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
<table width="20%" border="0" cellspacing="1" cellpadding="3">
<tr>
<td><h4> Student's Gender </h4></td>
<td>
<?php
$con=mysqli_connect("localhost", "root", "ew6wLoLOro", "result");
$sql=mysqli_query($con, "select gender_sl_no, gender_name from gender")
while($row=mysqli_fetch_array($sql))
{
echo '<table>
<input type="radio" name="sexbd" checked="checked"/>'.$row['gender_name'].'
</table>';
}
?>
</td>
</tr>
<tr>
<td> </td>
<td><input type ="submit" name="submit"/> </td>
</tr>
</table>
</form>
<?php
$con=mysqli_connect("localhost", "root", "ew6wLoLOro", "result");
$sexbd = $_POST['sexbd'];
if(isset($_POST['submit']))
{
echo $que="Insert into ajebaje VALUES(default,'$sexbd' )";
echo " ";
echo "Your Data Inserted";
$result = mysqli_query($con,$que);
}
?>
</body>
</html>
默认情况下,它应该在表单中发送sexbd: on
。
为了从表单发送值,需要在输入标签中定义value属性。 像这样:
echo '<table>
<input type="radio" name="sexbd" checked="checked" value='.$row['gender_name'].'/>'.$row['gender_name'].'
</table>';
呈现的表单将如下所示:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title> ajebaje </title>
</head>
<body>
<form name="form1" action= "/test" method="post">
<table width="20%" border="0" cellspacing="1" cellpadding="3">
<tr>
<td><h4> Student's Gender </h4></td>
<td>
<table>
<input type="radio" name="sexbd" checked="checked" value="Male"/> Male
</table>
<table>
<input type="radio" name="sexbd" checked="checked" value="Female"/> Female
</table>
</td>
</tr>
<tr>
<td> </td>
<td><input type ="submit" name="submit"/> </td>
</tr>
</table>
</form>
</body>
</html>
现在,提交的值将为sexbd: Male
或sexbd: Female
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