How can i pass Drop down values to sql database and also the check box for example if a user selects English
and maths
than the value inserted in to the database would be 1
or else the value would be 0
<form>
<p id="p1">Select Your Year</p>
<select id="year_sel">
<option value="blank"></option>
<option id="primary" value="primary">Primary</option>
<option value="1">Year one</option>
<option value="2">Year two</option>
<option value="3">Year Three</option>
</select>
<input type="checkbox" name="Math" value="Math">Math<br>
<input type="checkbox" name="English" value="English">English<br>
<input type="checkbox" name="HealthScience" value="HealthScience">Health Science<br>
<input class="sub_reg" type="submit" value="Register Subjects" />
</form>
答案 0 :(得分:2)
First, your <select id="year_sel">
needs a name attribute to post ->
<select id="year_sel" name="year_sel" >
Since you are using <form>
the default is get
, so you would get the selected value in $_GET
$year_sel = $_GET['year_sel'];
If you changed it to
<form method="post">
then you would get it in $_POST
$year_sel = $_POST['year_sel']
Second, checkboxes are only posted if checked, so you can use isset()
to set a value using a ternary -
$math = isset($_GET['Math']) ? 1 : 0;
$english = isset($_GET['English']) ? 1 : 0;
...[rest of your checkboxes]...
swap $_GET
/$_POST
like the select
$math = isset($_POST['Math']) ? 1 : 0;
$english = isset($_POST['English']) ? 1 : 0;
答案 1 :(得分:0)
<select id="year_sel" name="year_sel">
<option value="primary">Primary</option>
<option value="1">Year one</option>
<option value="2">Year two</option>
<option value="3">Year Three</option>
</select>
on your form action something.php file use this to get the select value
$language = $_POST["year_sel"]; // chose whetever your form method
establish the database connection please do refer some tutorials (if you don't know )
write the mysqli insert command like
$SQL = "INSERT INTO yourtable (language) VALUES ('$language')";
mysqli_real_escape_string($sql);
$result = mysqli_query($sql) or die (mysqli_error());
now you can insert this value in your mysql or any database table
this is not tested
答案 2 :(得分:0)
<form>
<p id="p1">Select Your Year</p>
<select id="year_sel" name="year_sel">
<option value="blank"></option>
<option id="primary" value="primary">Primary</option>
<option value="1">Year one</option>
<option value="2">Year two</option>
<option value="3">Year Three</option>
</select>
<?php
$variable = $_POST["year_sel"]; ?>
Should mention the HTML ELEMENT name to get the value.