Please look this link to see the code use by me
<pre>
http://jsfiddle.net/JaavS/
</pre>
<pre>
step1: select values in List A
step2: Move the values to list b
Step3:click save to database button
</pre>
After step3, the item in the list b values to be store in database. Can any one help me
Give your selects a name:
Select1:
<select name ="list1[]" id="list1" multiple="multiple" rows=2>
<option value=1>Option 1</option>
<option value=2>Option 2</option>
<option value=3>Option 3</option>
<option value=4>Option 4</option>
<option value=5>Option 5</option>
<option value=6>Option 6</option>
</select>
Select2:
<select name="list2[]" id="list2" multiple="multiple" rows=2>
</select>
In your php file:
<?php
//List 2
foreach ($_POST['list2'] as $selectedOption)
{
$query = "INSERT INTO table (option) VALUES ('$selectedOption')";
mysql_query($query) or die(mysql_error());
}
?>
Related
I am trying to send data from an HTML form to a MySQL database in phpmyadmin. I have a database named bhs2018 and a table called game 1. Here are the contents of that table.
Here is my form:
<form name="game" action="insert.php" method="post">
<p> <select id="player" name = 'player'>
<option value="b">B</option>
<option value="n">N</option>
<option value="a">A</option>
<option value="c">C</option>
<option value="m">M</option>
<option value="j">J</option>
<option value="ja">Ja</option>
</select>
<select id="what" name = 'what'>
<option value="shoton">Shot on Cage</option>
<option value="shotoff">Shot off Cage</option>
<option value="goal">Goal</option>
<option value="countergoal">Goal on Counter</option>
<option value="countershot">Shot on Counter</option>
<option value="assist">Assist</option>
<option value="block">Block</option>
<option value="steal">Steal</option>
<option value="turnover">Turnover</option>
<option value="drawn">Ejection Drawn</option>
<option value="ejected">Ejected</option>
</select>
<select id="where" name = 'where'>
<option value="set">Set</option>
<option value="navy">Navy</option>
<option value="leftwing">1/2 side past 5</option>
<option value="rightwing">4/5 side past 5</option>
<option value="point">Point/3</option>
<option value="lefttwo">1/2 side 2 meter</option>
<option value="righttwo">4/5 side 2 meter</option>
<option value="1">6 on 5 1</option>
<option value="2">6 on 5 2</option>
<option value="3">6 on 5 3</option>
<option value="4">6 on 5 4</option>
<option value="5">6 on 5 5</option>
<option value="6">6 on 5 6</option>
</select>
<select id="quarter" name = 'quarter'>
<option value="q1">Quarter 1</option>
<option value="q2">Quarter 2</option>
<option value="q3">Quarter 3</option>
<option value="q4">Quarter 4</option>
</select>
<select id="time" name = 'time'>
<option value="0:30">0:30</option>
<option value="1:00">1:00</option>
<option value="1:30">1:30</option>
<option value="2:00">2:00</option>
<option value="2:30">2:30</option>
<option value="3:00">3:00</option>
<option value="3:30">3:30</option>
<option value="4:00">4:00</option>
<option value="4:30">4:30</option>
<option value="5:00">5:00</option>
<option value="5:30">5:30</option>
<option value="6:00">6:00</option>
<option value="6:30">6:30</option>
<option value="7:00">7:00</option>
</select>
Notes: <input type="text" id = 'notes' name = 'notes'>
<button type="submit" onclick="save()"> Save </button> </p>
</form>
Whenever I click my "Save" button, the insert.php script loads. Instead of echoing something, it just shows the code. Here is insert.php.
<?php
$con = mysqli_connect('127.0.0.1','root','password'(my actual password is here);
if(!$con){
echo 'Not Connected to Server';
}
if (!mysqli_select_db($con,'bhs2018')){
echo 'Not Selected';
}
$Player = $_POST['player'];
$Quarter = $_POST['quarter'];
$Time = $_POST['time'];
$Where = $_POST['where'];
$Notes = $_POST['notes'];
$What = $_POST['what'];
$sql = "INSERT INTO game1 (player,quarter,time1,where1,notes,what) VALUES ('$Player', '$Quarter', '$Time', '$Where','$Notes','$What')";
if(!mysqli_query($con,$sql)){
echo'Not Inserted';
}
else{
echo 'Inserted';
}
header('refresh:2; url=index.html');
?>
What is looking wrong with my code? Why does it not run the php script? Thank you so much!
Check your database. Your PRIMARY KEY player should be an integer like int(30) or bigint(200) but not varchar. Create a new column for PRIMARY KEY something like player_id and shift player to next column. Also when primary key is an integer, there is no manual insertion option for inputting it's value in your form. So make your PRIMARY KEY is set to AUTO_INCREMENT in case you don't want to insert it's value manually.
I have a form with a dropdown list using <select> and <option> tags.
below that I have a textarea field.
The option ID's are the databases names, and I need to insert the textarea-input into the selected database.
Code:
HTML FORM:
<select class="form-control" id="selection" name="selection_name">
<option id="option1">First item</option>
<option id="option2">Second item</option>
</select>
PHP:
<?php
$dbToInsert = $_POST[''] // <-What do I insert here to get the selected option?
...
?>
You Should use "value" in option tag
<select name="vehicle">
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="mercedes">Mercedes</option>
<option value="audi">Audi</option>
</select>
In the above example, if you have chosen "Audi", on form submission you will get the value "audi" in $_POST["vehicle"]
Basically I'd like to use the choices which the user has selected from a different select tags, and in essence store these as variables which I can then use to query my database in SQL.
My HTML code is here:
<div id ="search_elements">
<img src="UniSelect.jpeg">
<select>
<option selected disabled>Select a university</option>
<option value="ucl">UCL</option>
<option value="kings">Kings College</option>
<option value="imperial">Imperial College</option>
<option value="lse">London School of Economics</option>
</select>
<img src="PriceSelect.jpeg">
<select>
<option selected disabled>Select a weekly rent price</option>
<option value="50">0-£50</option>
<option value="100"> £100-£150</option>
<option value="150">£150-200</option>
<option value="200"> £200+</option>
</select>
</div>
And the type of php i would be looking to use would be:
//$con=mysqli_connect("localhost","adam","YjM3ZTYwOTQ5OWRmYWZh","adam_...");
//if (mysqli_connect_errno())
// {
// echo "Failed to connect to MySQL: " . mysqli_connect_error();
// }
// Perform queries
//$sql=mysqli_query($con,"SELECT CONTENT FROM Blog WHERE ID = 01");
//$result=mysqli_fetch_assoc($sql);
//echo $result['CONTENT'];
//mysqli_close($con);
To make it clear once more, I want to use the different values which the user selects, upon clicking a search button, have these query results shown in a table.
This solution a little differs from yours because you have no provided your form, submit button, etc. but in general, after a few changes, it should work too:
<form method="post" action="">
<img src="UniSelect.jpeg">
<select name="university">
<option selected disabled>Select a university</option>
<option value="ucl">UCL</option>
<option value="kings">Kings College</option>
<option value="imperial">Imperial College</option>
<option value="lse">London School of Economics</option>
</select>
<img src="PriceSelect.jpeg">
<select name="rent_price">
<option selected disabled>Select a weekly rent price</option>
<option value="50">0-£50</option>
<option value="100"> £100-£150</option>
<option value="150">£150-200</option>
<option value="200"> £200+</option>
</select>
<input type="submit" value="Submit form">
</form>
And now, to get values of these (something like this and I recommend to place it above your form):
if (!empty($_POST)) {
// Checking connection in here
$university_id = mysqli_real_escape_string($con, $_POST['university']);
$rent_price = mysqli_real_escape_string($con, $_POST['rent_price']);
$sql = mysqli_query($con, "SELECT * FROM university WHERE name = '".$university_id."'");
$result = mysqli_fetch_assoc($sql);
// Same thing goes for rent price
}
I want a form to have multiple drop down selects that change based on the previous selection. This is how I want it to work.
<select id="format">
<option selected="selected" value="NULL">-- Select a product Format --</option>
<option value="PS3">PS3</option>
<option value="Xbox 360">Xbox 360</option>
<option value="Wii U">Wii U</option>
</select>
So if you select PS3 you are then faced with more options for PS3 product type
<select id="ps3-product-type">
<option selected="selected" value="NULL">-- Select a product PS3 product type --</option>
<option value="chargers">chargers</option>
<option value="controllers">controllers</option>
<option value="headsets">headsets</option>
</select>
So then you select headsets for example and you get another set of final options of products
<select id="ps3-headsets">
<option selected="selected" value="NULL">-- Select a PS3 headset --</option>
<option value="product-1">product 1</option>
<option value="product 2">product 2</option>
<option value="product 3">product 3</option>
</select>
How can I do this with jquery or PHP? Please bare in mind that I will also have selects for Xbox 360 and Wii U too.
UPDATE PROGRESS:
http://jsfiddle.net/maximus83/r7MN9/639/
Here is my HTML, I have got the first set of arrays working for selecting product type, But I cant get the 3rd box working, I need 3rd box to say the product, where do I go from here.
<select id="cat">
<option val="PS3">PS3</option>
<option val="Xbox360">Xbox360</option>
<option val="WiiU">WiiU</option>
<option val="Multiformat">Multiformat</option>
</select>
<select id="item">
</select>
<select id="product">
</select>
Here is my script
PS3=new Array('Headsets(PS3)','Controllers(PS3)','Chargers(PS3)','Cables(PS3)');
Xbox360=new Array('Headsets(360)','Chargers(360)');
WiiU=new Array('Controllers(WiiU)','Headsets(WiiU)');
Multiformat=new Array('Headsets(Multi)','Chairs(Multi)','Cables(Multi)');
populateSelect();
$(function() {
$('#cat').change(function(){
populateSelect();
});
});
function populateSelect(){
cat=$('#cat').val();
$('#item').html('');
eval(cat).forEach(function(t) {
$('#item').append('<option val="#item">'+t+'</option>');
});
}
This should do the trick for you. It's a previously answered thread on the exact same topic - how to populate a drop down based on the value in another. It takes advantage jQuery (And I've used it before myself as reference).
I have a problem with a dropdownlist created in Yii. I wanted to create a dropdown list with sample below:
<select id="con_id">
<option value="1">Con 1</option>
<option value="2">Con 2</option>
<option value="3">Con 3</option>
<option value="4">Con 4</option>
<option value="5">Con 5</option>
<option value="6">Con 6</option>
<option value="all">--All--</option>
</select>
Dropdown menu list are obtained from the database. Below is the sample code that I created in Yii format:
echo $form->dropDownList($model, '_conId', array_merge(CHtml::listData(Consultant::model()->findAll(), 'id', 'name'), array('client_id' => '--All--')));
However, the result of the dropdown list created in Yii become like this:
<select id="con_id">
<option value="">Con 1</option>
<option value="1">Con 2</option>
<option value="2">Con 3</option>
<option value="3">Con 4</option>
<option value="4">Con 5</option>
<option value="5">Con 6</option>
<option value="all">--All--</option>
</select>
The value on the option is not correctly assigned. Can anyone help me?
UPDATE:
Somehow I have found the solution:
echo $form->dropDownList($model, '_conId', array('all' => '--All--') + CHtml::listData(Consultant::model()->findAll(), 'id', 'name'));
There's an 'empty' option for CHtml::listData, where you can set the text for the 'empty' option.
Here's an example using your code:
<?php echo $form->dropDownList(
$model,
'_conId',
CHtml::listData(
Consultant::model()->findAll(),'id','name'), // list all data from model
array('empty'=>'--All--') // set empty data label
); ?>