I have a radio button with name and value , and I looked for many solutions but still i can't insert my radio button value on my database. My query is perfectly fine working, here is my code
<form action="" method="POST">
<label class="radio-inline"><input type="radio" name="gender" value="male">Male</label>
<label class="radio-inline"><input type="radio" name="gender" value="female">Female</label>
<input type="submit" class="btn btn-default" name="submit">
</form>
what is inserting on my database is the name not the value. so the word gender is the one who is inserting on my database.
I tired to change my gender field type into Varchar and Char but still not working.
if(isset($_POST['submit'])){
$gender = mysqli_real_escape_string($con, $_POST['gender']);
$sql = "INSERT INTO information_table (gender)
VALUES ('gender')";
Did you try like this??
<input type="radio" name="number" value="1" />
1
<input type="radio" name="number" value="2" />
2
<input type="radio" name="number" value="3" />
3
<input type="radio" name="number" value="4" />
4
<input type="radio" name="number" value="5" />
5
<?php
if (isset($_POST['submit'])){
$number = $_POST['number'];
mysql_query("INSERT INTO table_name(number) VALUES ('$number')");
}
?>
Related
I have several checkboxes that contain names (ids as referrence to database) - see code below. How can I select all checked values and add them to database (via MySQL) each row for each checked?
<input type="checkbox" value="1" name="names[]">John
<input type="checkbox" value="2" name="names[]">Peter
<input type="checkbox" value="3" name="names[]">Mike
<input type="checkbox" value="4" name="names[]">Kattie
<input type="submit" value="Send" name="send">
After clicking "send" with requiered checked names, the result in database should look like this (I selected John and Mike):
Id
1
3
(only selected ones)
How can I achieve that?
You need to wrap your inputs in a <form> element and give this form a method of post:
<form method="post">
<input type="checkbox" value="1" name="names[]" />John
<input type="checkbox" value="2" name="names[]" />Peter
<input type="checkbox" value="3" name="names[]" />Mike
<input type="checkbox" value="4" name="names[]" />Kattie
<input type="submit" value="Send" name="send" />
</form>
This will allow you to post submitted data from your form inputs to your PHP.
Note: If your HTML is in a different file (ie not in the same file as your form) you can add the action attribute to your form (eg: action="fileWithPHP.php")
Now you can access all checked checkboxes in your PHP using $_POST['names']. This will allow you to get your array of checked values. You can then use a foreach loop to loop through every value in your array:
<?php
if(isset($_POST['names'])) {
$names = $_POST['names'];
foreach($names as $name) {
echo "Add " . $name . " to db here<br />"; // add $name to db
}
}
?>
You can wrap the inputs around a <form> and send it to php and retrieve using $_GET or $_POST and update the database.
I have used POST method here.
HTML:
<form action="test.php" method="post">
<input type="checkbox" value="1" name="names[]">John
<input type="checkbox" value="2" name="names[]">Peter
<input type="checkbox" value="3" name="names[]">Mike
<input type="checkbox" value="4" name="names[]">Kattie
<input type="submit" value="Send" name="send">
</form>
PHP:
if(!empty($_POST['names'])) {
foreach($_POST['names'] as $check) {
echo $check; //instead of echo you can insert it to the database
}
}
I'm trying to get a value of a form in Wordpress to PHP. The form is like this and it is displaying fine in the preview:
<form action=".../name.php" method="get">
<input type="checkbox" name="form_ques_4" value=0 />
<input type="checkbox" name="form_ques_4" value=1 />
<input type="checkbox" name="form_ques_4" value=2 />
<input type="submit" name="formSubmit" value="submit" />
</form>
If the user selected option 2, the value is 1 and this will later be used as the input in a MySQL database. As I have read in other posts, I should get value with the php line.
$a = $_GET["form_ques_4"];
I have tested some other simple outputs for the .php and there is no problem with the "form action" of the wordpress. I also tried using single and double quotes for the "GET" with no result.
Try to change the names of your checkboxes, if you want a user multiple choice:
<form action=".../name.php" method="get">
<input type="checkbox" name="form_ques_1" value="0" />
<input type="checkbox" name="form_ques_2" value="1" />
<input type="checkbox" name="form_ques_3" value="2" />
<input type="submit" name="formSubmit" value="submit" />
</form>
otherwise, if you want the user makes only one choice use type="radio"
<form action=".../name.php" method="get">
<input type="radio" name="form_ques_4" value="0" />
<input type="radio" name="form_ques_4" value="1" />
<input type="radio" name="form_ques_4" value="2" />
<input type="submit" name="formSubmit" value="submit" />
</form>
EDIT
yes, as AZinkey says, you can also use
<form action=".../name.php" method="get">
<input type="checkbox" name="form_ques[]" value="0" />
<input type="checkbox" name="form_ques[]" value="1" />
<input type="checkbox" name="form_ques[]" value="2" />
<input type="submit" name="formSubmit" value="submit" />
</form>
then get the results in php
$checked = $_GET['form_ques'];
for($i=0; $i < count($checked); $i++){
echo $checked[$i] . "<br/>";
}
Quote your value attribute like value="0" and update name to "form_ques_4[]"
<input type="checkbox" name="form_ques_4[]" value="0" />
I have fetch data from database in $tableRe. Now I have to print values in textarea and also have to check the radio button.
Here is my code,
$sql = "select (address, gender) from stud table";
if($result=mysqli_query($conn,$sql)) {
while($row = mysqli_fetch_array($result)) {
$tableRe[]=$row;
}
}
<form>
Address : <br>
<textarea value="<?php echo $tableRe[0]['address']; ?>"></textarea><br>
Gender : <br>
<input type="radio" value="Male">Male
<input type="radio" value="Female">Female <br>
<input type="submit" value="Save">
</form>
Please help me regarding this. Thanks in advance.
You need to apply condition on checked HTML attribute.
Try this:
<form>
Address : <br>
<textarea><?php echo $tableRe[0]['address']; ?></textarea> <br/>
Gender : <br>
<input type="radio" value="Male" <?php echo $tableRe[0]['gender'] == 'Male' ? 'checked' : ''; ?> >Male
<input type="radio" value="Female" <?php echo $tableRe[0]['gender'] == 'Female' ? 'checked' : ''; ?>>Female <br>
<input type="submit" value="Save">
</form>
Value has to be placed between the openning and closing tags :
<texterea name="whatever">Textarea value goes here</textarea>
To set a radio/checkbox as selected/checked, you need to add to it a "checkded" attribute :
<!-- HTML4 -->
<input type="radio" name="whatever" value="value1" checked="checked" /> Label 1
<input type="radio" name="whatever" value="value2" /> Label 2
<input type="radio" name="whatever" value="value3" /> Label 3
<!-- HTML5 -->
<input type="radio" name="whatever" value="value1" checked /> Label 1
<input type="radio" name="whatever" value="value2" /> Label 2
<input type="radio" name="whatever" value="value3" /> Label 3
I'm getting a strange error when I try to submit user-generated data to a database via PHP commands. When I hit the submit button below, instead of the PHP page running its' function I am presented with a display of the raw code on my browser. I have a command at the bottom of my HTML page that looks like this:
<form action="insert.php" method="post">
<input type="submit">
</form>
So that when the user hits the submit button, the PHP file insert.php (detailed below) is called to input the answers onto a database, separating each answer into it's own field.
Here is the code I'm working with:
<?php
$con=mysqli_connect("host","username","password","database");
// Check connection
if (mysqli_connect())
{
echo "Failed to connect to MySQL: " . mysqli_errno();
}
$sql="INSERT INTO Persons (Name, Serif, Width, Height, Spacing, Weight)
VALUES
('$_POST[answer]','$_POST[answer]','$_POST[answer]','$_POST[answer]','$_POST[answer]','$_POST[answer]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
Right now, the questions are in a and not a (is there a functional difference in this case?). They look like:
<form class="testAns" id="widthAns">
<input type="radio" name="answer" value="skinny">-25%
<input type="radio" name="answer" value="skinny">-10%
<input type="radio" name="answer" value="mid">normal
<input type="radio" name="answer" value="fat">+10%
<input type="radio" name="answer" value="fat">+25%
</form>
<form class="testAns" id="spaceAns">
<input type="radio" name="answer" value="small">-25%
<input type="radio" name="answer" value="small">-10%
<input type="radio" name="answer" value="mid">normal
<input type="radio" name="answer" value="wide">+10%
<input type="radio" name="answer" value="wide">+25%
</form>
<form class="testAns" id="weightAns">
<input type="radio" name="wanswer" value="light">-25%
<input type="radio" name="answer" value="light">-10%
<input type="radio" name="answer" value="mid">normal
<input type="radio" name="answer" value="heavy">+10%
<input type="radio" name="answer" value="heavy">+25%
</form>
<form method="post" action="insert.php" class="testAns" id="heightAns">
<input type="radio" name="answer" value="short">-25%
<input type="radio" name="answer" value="short">-10%
<input type="radio" name="answer" value="mid">normal
<input type="radio" name="answer" value="tall">+10%
<input type="radio" name="answer" value="tall">+25%
</form>
The important part is for the "value" associated with each button to be logged into the database. For example, if a user selects "+10%" I want be able to log the word "heavy".And then there are two text input fields:
<form id="intro">
City: <input type="text" name="answer"><br>
Why you are using this tool:<input type="text" name="answer">
</form>
So for these text fields I need the user input logged as the answer.
I see you got the PHP thing fixed.
Now you need to fill your form with data. This:
<form action="insert.php" method="post">
<input type="submit">
</form>
sends only the submit value. You need to add input fields inside that form tag, otherwise, nothing else will get sent. So, since you're sending an answer array, you should add those (adding them as text fields, as an example):
<form action="insert.php" method="post">
<input type="text" name="answer[]" />
<input type="text" name="answer[]" />
etc...
<input type="submit" />
</form>
And make sure you filter all user inputs before writing anything into the database, as otherwise my buddy Bobby Tables might come to visit you.
Make sure in your XAMPP Control Panel that Apache and MySQL are running. Then check if your input fields are inside the <form action='insert.php' method='POST'> input fields </form>
Your HTML code would look like this:
<html>
<body>
<form action='insert.php' method='POST'>
<table>
<tr><td>Width: </td><td>
<input type="radio" name="width" value="skinny">-25%
<input type="radio" name="width" value="skinny">-10%
<input type="radio" name="width" value="mid">normal
<input type="radio" name="width" value="fat">+10%
<input type="radio" name="width" value="fat">+25%
</td></tr>
<tr><td>Spacing: </td><td>
<input type="radio" name="spacing" value="small">-25%
<input type="radio" name="spacing" value="small">-10%
<input type="radio" name="spacing" value="mid">normal
<input type="radio" name="spacing" value="wide">+10%
<input type="radio" name="spacing" value="wide">+25%
</td></tr>
<tr><td>Weight: </td><td>
<input type="radio" name="weight" value="light">-25%
<input type="radio" name="weight" value="light">-10%
<input type="radio" name="weight" value="mid">normal
<input type="radio" name="weight" value="heavy">+10%
<input type="radio" name="weight" value="heavy">+25%
</td></tr>
<tr><td>Height: </td><td>
<input type="radio" name="height" value="short">-25%
<input type="radio" name="height" value="short">-10%
<input type="radio" name="height" value="mid">normal
<input type="radio" name="height" value="tall">+10%
<input type="radio" name="height" value="tall">+25%
</td></tr>
<tr><td>City: </td><td><input type="text" name="city"></td></tr>
<tr><td>Why you are using this tool: </td><td><input type="text" name="tool"></td></tr>
<tr><td></td><td><input type='submit'></td></tr>
</table>
</form>
</body>
</html>
What are you using in creating your php files? Dreamweaver? Notepad? Try this: SAVE AS your file, Save As Type: All Files and name it insert.php.
<?php
$con=mysqli_connect("localhost","YourUsername","YourPassword(if any)","NameOfYourDatabase");
// Check connection
if (mysqli_connect())
{
echo "Failed to connect to MySQL: " . mysqli_errno();
}
$width=$_POST['width'];
$spacing=$_POST['spacing'];
$weight=$_POST['weight'];
$height=$_POST['height'];
$city=mysqli_real_escape_string($con,$_POST['city']);
$tool=mysqli_real_escape_string($con,$_POST['tool']);
/* REAL ESCAPE STRING WOULD PREVENT A BIT OF SQL INJECTION */
$sql="INSERT INTO Persons (Name, Serif, Width, Height, Spacing, Weight)
VALUES
('$city','$tool','$width','$height','$spacing','$weight')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
i want insert this form value to datanase :
<input type="checkbox" name="brand1" id="brand1" value="1"> <label for="brand1">Brand 1</label>
<input type="checkbox" name="brand2" id="brand2" value="1"> <label for="brand2">Brand 2</label>
<input type="checkbox" name="brand3" id="brand3" value="1"> <label for="brand3">Brand 3</label>
<input type="checkbox" name="brand4" id="brand4" value="1"> <label for="brand4">Brand 4</label>
<input type="checkbox" name="brand5" id="brand5" value="1"> <label for="brand5">Brand 5</label>
these text box are get by php from a table in database and may be Variable
i want insert to database by this format
if brand 1 are checked $brand="1,";
and Finally like this :
insert($name,$brands); and $brands = "1,2,3,4,5,";
if write this by if and while but it doesn't work because if insert run in while {} Five times insert Done and if insert run out of while {} , $brand = "5,"
thanks for your help or idea for this problem
it's mean :
<form method="post" action="#">
<?php
$result = $db->getall(brands);
if(!empty($result)) {
while ( list($key,$val)=each($result) ) {
$brand_id = stripslashes($val["id"]);
$brand_name = stripslashes($val["name"]);
?>
<input type="checkbox" name="brand<?php print"$brand_id"; ?>" value="1" style="cursor:pointer;"><label for="brand<?php print"$brand_id"; ?>" style="cursor:pointer;"> <?php print"$brand_name"; ?></label>
<?php }} ?>
Source Output:
<input type="checkbox" name="brand1" value="1"> <label for="brand1">Brand Name 1</label>
<input type="checkbox" name="brand2" value="1"> <label for="brand2">Brand Name 2</label>
<input type="checkbox" name="brand3" value="1"> <label for="brand3">Brand Name 3</label>
<input type="checkbox" name="brand4" value="1"> <label for="brand4">Brand Name 4</label>
<input type="checkbox" name="brand5" value="1"> <label for="brand5">Brand Name 5</label>
<input type="submit" value="Submit" />
</form>
when submit form , insert source is :
<?php
$result = $db->getall(brands);
if(!empty($result)) {
while ( list($key,$val)=each($result) ) {
$brand_id = brand.stripslashes($val["id"]);
$brand_name = stripslashes($val["name"]);
$brand_ids = "brand.$brand_id";
if($$brand_ids==1) {$brands="$brandid,"}
}} ?>
$db->add_submenu("$brands");
You should change the name of your checkboxes to brand[]. It will give you an array once submitted at $_POST['brand']
Ex.
<input type="checkbox" name="brand[]" value="1" ... />
<input type="checkbox" name="brand[]" value="2" ... />
<input type="checkbox" name="brand[]" value="3" ... />
<input type="checkbox" name="brand[]" value="4" ... />
<input type="checkbox" name="brand[]" value="5" ... />
on the other side you can either do something like the following:
// this will return '1, 2, 3, 4, 5' when all are selected.
$index = implode(", ", $_POST['brand']);
and at that point you will have the brands in comma delimited form.