I have the following form:
<form name='progObj_form' method='POST' enctype='multipart/form-data' action='processpage.php'>
<select name='manageObj[]' id='objectives' multiple="multiple">
<option value=0>there are no objectives for this program</option>
</select><br />
<a href='#nogo' onclick="delItem(objectives,0,'objEditBtn')" class='shiftOpt'>delete selected</a><br />
<input name='newObjective' type='text' id='newObjective'/>
<input name='addNew' type='button' onclick="AddItem(newObjective.value,6,'objectives','objEditBtn');" value='add objective'/>
<input name="passProgID" type="hidden" value="1" /><br />
<input name="objectiveEdit" id="objEditBtn" type="submit" value="save changes" disabled=disabled/>
</form>
that allows data (objectives in this case) to be added and deleted from a list box. That all works well but for some reason the updated listbox values aren't being passed to the process page.
I'm catching the data like so (simplified):
if (isset($_POST['objectiveEdit'])){
$progID=$_POST['passProgID'];
for ($v=1;$v<count($_POST['manageObj']);$v++){
$value=$_POST['manageObj'][$v];
$sqlObj="INSERT INTO progObjective (progID,objective,objectiveOrder) VALUES ($progID,$value,$v)";
$result = mssql_query($sqlObj,$linkProbation) or die('Query failed: '.$sqlObj);
}//end for ($a=0;$a<count($_POST['manageObj']);$a++)
$objMsg=print_r($_POST['manageObj']).$sqlObj;
}//end if (isset($_POST['objectiveEdit'])
For $objMsg, I get a response of 1 and the array doesn't print because ostensibly, it's empty which means that it also doesn't enter the for loop.
I can include the javascript as well but started with just this for simplicity since I'm probably just overlooking something obvious?!
Option elements are never sent to the server. Only the value of the selected option will be sent.
In your case, something like manageObj=x will be sent to the server, where x is the value of the option element that is selected by the user. It is possible that you misunderstand the [] when it's used in a name attribute.
You should try to find a different method if you want to send the objectives created by the user to the server. You could store the data in hidden inputs for example.
So I finally figured it out! There's no array because there are no selections made! I updated the submit button to call a SELECT ALL function and now it's all good.
<input name="objectiveEdit" id="objEditBtn" type="submit" value="save changes" onclick="selectAll('objectives',true)" />
name='manageObj[]' - shouldn't that be name='manageObj' ?
Related
I have an input type text box as follows
<input type="text" name="deleteprofileconfirmation" id="deleteprofileconfirmation" class="editprofileinput">
Delete Account
I need to pass the value entered in the input type text to deleteaccount.php
I can do with help of jquery, no problem, i need a pure php solution...
I tried using sessions, but problem is how to read the value in input type when link is clicked.. $_POST is also not working...
i cannot use form because this is a form in another form so html5 is not allowing nested forms, sorry should have mentioned that earlier
the following is not working on deleteaccount.php
if (isset($_POST['deleteprofilebutton']))
{
$delete_profile = strtolower($_POST['deleteprofileconfirmation']);
}
Make your link as
href="../controllers/deleteaccount.php?id=$ID_VALUE"
and update the POST to GET
if (isset($_GET['id']))
{
$delete_profile = strtolower($_GET['id']);
}
That would be GET now.
Make sure users with no privileges can hit this url and delete the profiles. Do check the user rights before processing the delete operation.
You can do this using form
<form action="../controllers/deleteaccount.php" method="post">
<input type="text" name="deleteprofileconfirmation" id="deleteprofileconfirmation" class="editprofileinput">
<input type="submit" class="deleteprofilebutton" name="deleteprofilebutton" id="deleteprofilebutton" value="Delete Account">
</form>
You could also give each one a unique name and just check the $_POST for the existence of that input.
<input type="submit" name="deleteprofileconfirmation" value="Delete" />
<input type="submit" name="submit" value="Submit" />
And in the code:
if (isset($_POST['deleteprofileconfirmation'])) {
//delete action
} else if (isset($_POST['submit'])) {
//submit action
} else {
//no button pressed
}
I want to create a checkbox and check whether checkbox is checked or not in another PHP page so I created form like :
<form action="heartbeat.php" method="post">
<input type="checkbox" name="keepme">
<input type="submit" value="log in">
</form>
and then I tried to extract this input on heartbeat.php like:
<?php
/*
* A PHP file for laying down a heartbeat JavaScript call.
*/
if(!isset($_POST["keepme"])){
$auto_logout = 10;
}
else{
$auto_logout = 1000;
}
?>
but it never gets $_POST["keepme"] value. Any idea??
If you don't pass value to checkbox tag, the POST array will be empty. You need to do something like that
<input type="checkbox" name="keepme" value='1'>
Make Sure you have submit button
If(isset($_POST['submit-btn'])){
echo $_POST['keepme'];
}
This is my second code but the problem is I have 3 queries. So it only returns the last product_id when i Click update it always return product_id=3, but i want update the product_id=2
<form action="update_qty.php" method="POST">
<?php while($getorder = mysqli_fetch_array($order)){ ?>
<input type="hidden" value="<?=$getorder['price']?>" name="actual_price">
<input type="hidden" value="<?=$getorder['product_id']?>" name="product">
<input type="text" value="<?=$getorder['qty']?>" name="qty" size="1" style="text-align:center">
<input type="submit" value="update" name="update">
<?php } ?>
</form>
Your problem is that the PHP is server side and you need something client side to read the value of the text box. You would need a page refresh to pass the text field value to the server so it could write it to the url in the anchor tag. Which is what the form submit would do, but as it would have submitted the value already the anchor tag would be pointless
To do it without a page refresh use Javascript. It would be easy to do with jQuery. You could add an event that writes whatever is entered in the text box the the anchor tags href as it is typed.
I'll do something more like this.
One form per product.In your case when you submit the form the qty value will always be the las found.
<?php while($getorder = mysqli_fetch_array($order)){ ?>
<form action="update_qty.php" method="POST">
<input type="hidden" value="<?=$getorder['price']?>" name="actual_price">
<input type="hidden" value="<?=$getorder['product_id']?>" name="product">
<input type="text" value="<?=$getorder['qty']?>" name="qty" size="1" style="text-align:center">
<input type="submit" value="update" name="update">
</form>
<?php } ?>
You can add more information like this
update
You can not get all values as like that because input name overwrite in every loop iteration.
For multiple values you can try in two ways like:
<?php
while($getorder = mysqli_fetch_array($order)){
$newArr[] = $getorder['price']."~".$getorder['product_id']."~". $ getorder['qty'];
} //while end
?>
<input type="hidden" name="allinputs" value="<?=$newArr?>">
Input field outside the loop.
In php explode array value with ~ and get the all values.
Other solution is that
Your input field name must be change like:
<?php while($getorder = mysqli_fetch_array($order)){ ?>
<input type="hidden" value="<?=$getorder['price']?>" name="actual_price_<?=$getorder['product_id']?>">
<?php } ?>
Change field name in every iteration.
In current scenario either you need three different buttons or the best solution to use AJAX request .
update
On update_qty.php u can use like this
<?php echo $_GET['product_id'];?>
I have a list of names and some buttons with product names. When one of the buttons is clicked the information of the list is sent to a PHP script, but I can't hit the submit button to send its value. How is it done?
I boiled my code down to the following:
The sending page:
<html>
<form action="buy.php" method="post">
<select name="name">
<option>John</option>
<option>Henry</option>
<select>
<input id='submit' type='submit' name='Tea' value='Tea'>
<input id='submit' type='submit' name='Coffee' value='Coffee'>
</form>
</html>
The receiving page: buy.php
<?php
$name = $_POST['name'];
$purchase = $_POST['submit'];
//here some SQL database magic happens
?>
Everything except sending the submit button value works flawlessly.
The button names are not submit, so the php $_POST['submit'] value is not set. As in isset($_POST['submit']) evaluates to false.
<html>
<form action="" method="post">
<input type="hidden" name="action" value="submit" />
<select name="name">
<option>John</option>
<option>Henry</option>
<select>
<!--
make sure all html elements that have an ID are unique and name the buttons submit
-->
<input id="tea-submit" type="submit" name="submit" value="Tea">
<input id="coffee-submit" type="submit" name="submit" value="Coffee">
</form>
</html>
<?php
if (isset($_POST['action'])) {
echo '<br />The ' . $_POST['submit'] . ' submit button was pressed<br />';
}
?>
Use this instead:
<input id='tea-submit' type='submit' name = 'submit' value = 'Tea'>
<input id='coffee-submit' type='submit' name = 'submit' value = 'Coffee'>
The initial post mentioned buttons. You can also replace the input tags with buttons.
<button type="submit" name="product" value="Tea">Tea</button>
<button type="submit" name="product" value="Coffee">Coffee</button>
The name and value attributes are required to submit the value when the form is submitted (the id attribute is not necessary in this case). The attribute type=submit specifies that clicking on this button causes the form to be submitted.
When the server is handling the submitted form, $_POST['product'] will contain the value "Tea" or "Coffee" depending on which button was clicked.
If you want you can also require the user to confirm before submitting the form (useful when you are implementing a delete button for example).
<button type="submit" name="product" value="Tea" onclick="return confirm('Are you sure you want tea?');">Tea</button>
<button type="submit" name="product" value="Coffee" onclick="return confirm('Are you sure you want coffee?');">Coffee</button>
To start, using the same ID twice is not a good idea. ID's should be unique, if you need to style elements you should use a class to apply CSS instead.
At last, you defined the name of your submit button as Tea and Coffee, but in your PHP you are using submit as index. your index should have been $_POST['Tea'] for example. that would require you to check for it being set as it only sends one , you can do that with isset().
Buy anyway , user4035 just beat me to it , his code will "fix" this for you.
Like the others said, you probably missunderstood the idea of a unique id. All I have to add is, that I do not like the idea of using "value" as the identifying property here, as it may change over time (i.e. if you want to provide multiple languages).
<input id='submit_tea' type='submit' name = 'submit_tea' value = 'Tea' />
<input id='submit_coffee' type='submit' name = 'submit_coffee' value = 'Coffee' />
and in your php script
if( array_key_exists( 'submit_tea', $_POST ) )
{
// handle tea
}
if( array_key_exists( 'submit_coffee', $_POST ) )
{
// handle coffee
}
Additionally, you can add something like if( 'POST' == $_SERVER[ 'REQUEST_METHOD' ] ) if you want to check if data was acctually posted.
You can maintain your html as it is but use this php code
<?php
$name = $_POST['name'];
$purchase1 = $_POST['Tea'];
$purchase2 =$_POST['Coffee'];
?>
You could use something like this to give your button a value:
<?php
if (isset($_POST['submit'])) {
$aSubmitVal = array_keys($_POST['submit'])[0];
echo 'The button value is: ' . $aSubmitVal;
}
?>
<form action="/" method="post">
<input id="someId" type="submit" name="submit[SomeValue]" value="Button name">
</form>
This will give you the string "SomeValue" as a result
https://i.imgur.com/28gr7Uy.gif
I'm trying to get a website working. What I have are basically two images displayed (random, taken out of a mySQL database). What I need to do is (when the user clicks one of the images) the following:
Update the page, passing the info about the selected image (submit form);
Add one piece of data to the database (upvote the image)
I need to use $_POST to pass an array of values to the next page. So I thought:
<form name="input" action="the_page.php" method="POST">
<input type="image"
name="img"
src="image.png"
value ="dat1[\"data1\",\"data2\",\"data3\"]">
<!-- If value must be a single string, I'll use hidden inputs-->
</form>
<form name="input" action="the_page.php" method="POST">
<input type="image"
name="img"
src="image2.png"
value ="dat2[\"data1\",\"data2\",\"data3\"]">
</form>
Then I can upvote the selected image on the mySQL database with a little php upvote() function that updates the record. The upvoting process is done when the new page is loaded. From this, I have a couple questions:
I'm guessing the images will act as buttons, right? (They are supposed to submit the form, hence refreshing the page). If not, how can I achieve this? I'm unable to do it with a link (since I can't add the values to it). Maybe a javascript function? But I don't know how to submit the form that way either...
Once the page is reloaded, does it mean that only the data from one form has been submited, so I can retrieve the data by simply calling the PHP variable $_POST['img'] and get an array back?
EDIT: I now managed to get everything working, slightly similar to what I proposed initially. Thanks for the AJAX suggestion though, since it was what helped me solve it (looked up AJAX tutorials, found solution).
Here's my solution:
<?php
echo "<form name=\"input\" action=\"F2F.php\" method=\"POST\">";
echo "<input type=\"hidden\" name =\"table\" value=\"".$table1."\">";
echo "<input type=\"image\" name=\"nom\" src=\"".$IMG_Route1."\" value =\"".$Nom_base1."\" border=\"0\">";
echo "</form>";
?>
(where the image goes)
and then, on the header:
<?php
if ($_POST['nom']||$_POST['nom_x']){
if (!$_POST['nom']){
echo 'Could not retrieve name. $_POST[\'nom_x\'] = '.$_POST['nom_x']. mysql_error();
exit;
}
if (!$_POST['table']){
echo 'Could not retrieve table. $_POST[\'table\'] = '.$_POST['table']. mysql_error();
exit;
}
upvote($_POST['table'],$_POST['nom']);
}
?>
You can use one form and a set of radio buttons to simplify things a bit. Clicking on the label will toggle the radio button. You can use commas to separate multiple values for each checkbox, which you can then abstract later on (see below)
<form name="input" action="the_page.php" method="POST">
<ul>
<li>
<label>
<img src="whatever.jpg" />
<input type="radio" name="selectedImage" id="img1" value="12,16,19" />
</label>
</li>
<li>
<label>
<img src="whatever2.jpg" />
<input type="radio" name="selectedImage" id="img2" value="12,16,19" />
</label>
</li>
</ul>
</form>
You can detect when the radio button is selected by adding a listener for the change event, then submit the form.
$('input[name="selectedImage"]').change(function() {
$('form[name="input"]').submit();
});
To abstract the multiple values, you can then explode the form result with PHP, which will return an array of the values.
$selectedImageValues = array();
$selectedImageValues = explode(",", $_POST['selectedImage']);
From there you can pull the different values out and save the data to the database.