I am working on a PHP project - I had one form post a date to another form
I made some changes (although none to the input in question)
Now all other inputs are updated with their Posted values, except the date
If I manually set the date in HTML it works:
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="2009-01-01"></div>
If I set it to the following, it doesn't:
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="<?php echo (isset($DateCourse))?$DateCourse:'';?>"></div>
The below:
$DateCourse = ($_POST["DateCourse"]);
var_dump($_POST["DateCourse"]);
var_dump($DateCourse);
Returns:
string(10) "2019-01-05" - means the post value is set
Notice: Undefined variable: DateCourse in /home/bitecons/bts.biteconsulting.co.za/v2/editccr.php on line 119 - how can it be undefined, I just defined it
NULL
What on earth am I doing wrong? Apart from using PHP :P
Flow as requested:
Records.php:
This is the function to prepopulate my posted fields:
function Prefill(x) {
TabletoEdit = x.closest('table').id;
SelectedRow = x.rowIndex;
document.getElementById("EntryEditing").value = x.cells[19].innerHTML;
document.getElementById("DateCourse").value = x.cells[0].innerHTML;
document.forms["records"].submit();
}
Then I also have:
<form action="editrec" method="post" id="records">
<input type='hidden' name='Period' id='Period' />
<input type='hidden' name='Month' id='Month' />
<input type='hidden' name='res' id='res' />
<input type='hidden' name='CustName' id='CustName' />
<input type='hidden' name='DateCourse' id='DateCourse' />
</form>
The Prefill gets called, then submits the form
I have tracked and DateCourse has data, but when getting to the other form, it "disappears":
if(!empty($_POST)) {
$DateCourse = ($_POST["DateCourse"]);
$CustName = ($_POST["CustName"]);
}
For example, CustName is filled in, but not DateCourse?
Side question:
Would this return true if another post var is not set (unrelated to this one):
if(!empty($_POST))
I think You use wrong Code
you Must Submit First Form And Then Use $DateCourse this In another Form in POSTBACK
One of the best way is to define $DateCourse too like
<?php
$DateCourse = "";
if(!empty($_POST["DateCourse"])) {
$DateCourse = ($_POST["DateCourse"]);
}
?>
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="<?php echo $DateCourse;?>"></div>
Okay, apologies folks, but it may help others in the future.
I had a function call to an old function - this failed, causing my variable to never get defined... I knew it was something stupid, but sometimes one needs a sound board...
Related
im having this variable:
var setValuey = function(x) {
document.getElementById('atype').value= x;
}
this refers to a clicked element:
<input type="hidden" name="type" value="" id="atype"><div class="person"><img src="images/voksenknap.png" onClick="setValuey('voksen')">
the problem is now that the set value (which happens when you click the element needs to be send to a PHP form on another site called resu.php which has this form:
form method="get" action="koeb.php">
<fieldset><legend>Billet</legend> <table border="0"><tr> <td>Type</td><td> <input type="text" name="zone" value="<?php echo (?????);?>" readonly id="atype" style="margin-left: 20px"> </td> </tr>
how do i get the set value into the value field ? whenever i am writing something in the value="<?php echo (?????);?>" all i get is a text of an undefined variable or the text setValuey:
<br /><b>Notice</b>: Undefined variable: setValuey in <b>C:\xampp\htdocs\interaktion\01-company\resu.php</b> on line <b>18</b><br />
that is what is being written in the value field for example.
hope my description is understandable. i really need help with this.
The value is put into a form input. On the server, you read that with $_POST['type'], so it should be:
value="<?php echo $_POST['type']; ?>"
I have strange problem with my search form. After I enter keyword and do the search request I get empty parameter value.
For example I type in the search field the word "something"
I see an empty value:
search.php?keyword=
After this I enter the keyword "else" and I recieve:
search.php?keyword=something instead of search.php?keyword=else
They somehow appear with "one step back"
I was trying to debug with print_r and var_dump but I only can print some values that does not explain my problem.
Am I missing something very trivial?
Here is what I have:
My class function:
public function show_search_result() {
$this->search_keywords = strip_tags($_GET['keyword']);
$this->_db->query("SELECT * from posts WHERE post_title LIKE '%$this->search_keywords%' OR post_content LIKE '%$this->search_keywords%' LIMIT 100");
$this->rows_results_found = $this->_db->resultset();
}
And my form:
<form action="search.php?keyword=<?php
if (isset($search_results->rows_results_found) && isset($_POST['search_requested'])) {
print strip_tags($_POST['search_keywords']);
}
?>" method="post">
<input type="hidden" name="search_requested">
<input type="text" name="search_keywords" value="<?php
if (isset($search_results->rows_results_found) && isset($_POST['search_requested'])) {
print strip_tags($_POST['search_keywords']);
}
?>"><input type="submit" value="Search">
</form>
<form action="" method=get>
<input type=text id=se>
<?php
if($_GET != null){
$sekw = $_GET ['se'];
$sql = //the query like='$sekw' limit=100;
?>
<input type=submit>
</form>
A simple code.
Your problem is when you send the form, it does save the keywords until the second send.
change your method from post to get. also i would advice you to use a framework for easy and fast coding. some include symfony2, laravel
I have a php variable "echo $id". Now I want to use the $_POST method to post the variable. I just want to know how to do this for a variable because $_POST[$id] does not work?
I think you are misunderstanding a basic concept here.
The $_POST super global is used to receive input (in the form of a POST request) from the user. While it is possible to set variables in it, you shouldn't.
Your question does not make sense. If you have an HTML form:
<form action="" method="post">
<input type="text" name="something" />
<input type="submit" value="Submit" />
</form>
Then you get the variable $_POST['something'] with whatever the user typed in the text box.
On its own, $_POST is just a variable like any other. You can assign to it $_POST['test'] = 123;, you can delete from it unset($_POST['test']);, you can even make it something other than an array $_POST = "Hello, world";, it just happens to be pre-populated with form data, if any.
With the method $_POST you must be posting to something.
My suggestion to you is to create a form, then have the form going to the file you wish to post to:
So something like this:
echo '<form action = "fileToPostTo.php" method = "post">
<input type = "text" hidden value = "'.$id.'" />
</form>';
And then submit the form when the document loads through jquery or javascript.
You can do it by $_POST['id'] = $id (then You will have it in $_POST['id'] variable (but You shouldn't do it :P).
Or You can send $id by form. Like example:
<form action="/pageToPOST.php" method="post">
<input type="text" value="<?=$id ?>" name="id" />
<input type="submit" name="" value="submit it!" />
</form>
And You'll have $_POST['id'] on http://yourdomainname.com/pageToPOST.php page
you can get and pass the value without page load and form.
<input type="text" name="something" id="something" />
<input type="button" value="ok" onclick="value();"/>
function value()
{
var something=$("#something").val();
var dataparam="oper=show&something="+something;
$.ajax({
type:"post",
url:"yourphpname.pnp",//this is very important.
data:dataparam,
success:function(data)
{
alert(data);
}
});
}
$oper =(isset( $_REQUEST['oper'])) ? $_REQUEST['oper'] : '';
if($oper == "show" and $oper != '')
{
$something=$_REQUEST['something']
echo $something;
}
what you want to do is assign a value submitted to your script using the POST method, to your $id variable. Something like:
$id = $_POST['id'];
I can't set a variable from a post array.
I have a simple form with a hidden field in it:
<input name="sid" type="hidden" id="sid" value="<?=$sid?>">
This hidden field gets sent off to a second file (exec.php) where I have the following code:
$sid = $_POST['sid'];
For some reason, when trying to set $sid, it gets a NULL value. For haha's, I ran the following:
foreach($_POST as $var => $value)
{
echo $var . ' : ' . $value . "<br>";
}
This provided a correct value of 1938 for sid. I've looked at this for 3 hours and can't find what is happening. I expect something extremely stupid...any thoughts?
Here is the form on enter.php
<form name="form1" method="post" action="exec.php">
<input name="sid" type="hidden" id="sid" value="<? echo($sid); ?>">
<input name="ticket_totals" type="hidden" id="ticket_totals" value="<?=$ticket_totals?>">
<input name="emp" type="hidden" id="emp" value="<?=$emp?>">
<input name="submit" type="submit" id="submit" value="Submit">
<input type="submit" name="submit" id="submit" value="Close">
</form>
Here is the POST output on exec.php:
type : Other
ticket_totals : 0
emp : 105
sid : 1939
submit : Submit
Okay - this was poor syntax on my part but now I'm curious as to why.
I left out quotation marks - the solution is as simple as this:
$sid = $_POST["sid"]
Now it works like a champ.
Any takers on why? I'd guess there is a setting in the php.ini that requires the quotes. Strangely enough, I have other variables called from the POST array that i'm not using quotes for and they're working fine...
Use Console in FireBug to inspect the POST request to see what is the sid value that is being sent.
If the sid value in request is ok, use var_dump($_POST["sid"]); to see the results on the server.
EDIT: it's considered good PHP style to use the quotes when accessing the associative array because quote-less keys are indistinguishable from constants:
define('myVar',3);
echo $array[myVar]; // retrieves $array[3], not $array['myVar'];
Try to echo the $sid instead of the <?=:
// Change that
<input name="sid" type="hidden" id="sid" value="<?=$sid?>">
// With that
<input name="sid" type="hidden" id="sid" value="<?php echo $sid; ?>">
also for the test time try to change the input type from hidden to text in order to be 100% sure the $sid contains a value.
Using quotes for associative array keys is mandatory, and while it may work without them, it's incorrect and erratic behavior is expected.
I had this same problem, trying to use $_POST[sid] as a variable. I'm am thinking that "sid" is a reserved or restricted variable name, because I changed my variable to $_POST[snid] and it worked just fine. This was my code
$sid = $_POST[sid];
$recipient = "($sid) ($_POST[sid])";
if ($_POST[sid] > 0)
{
$recipient = "It Worked";
}
print $recipient;
When I posted "&sid=15", the result was:
() (15)
Unbelievable. Impossible, right? All I did was change from using "sid" as the index to "snid", and it worked no problem.
So, don't ever use $_POST[sid].
I am just trying to learn PHP and want to get the value of the textbox using $_post function, but its not working. I am using wamp 2.1 and the code is simple as below
<form method="POST" action="c:/wamp/www/test/try.php">
<input type="text" name="nco" size="1" maxlength="1" tabindex="1" value="2">
<input
tabindex="2" name="submitnoofcompanies" value="GO"
type="submit">
</form>
<?php
if (!isset($_POST['nco']))
{
$_POST['nco'] = "undefine";
}
$no=$_POST['nco'];
print($no);
However in no way I get the value of the textbox printed, it just prints undefined, please help me out.
You first assigned the word "undefine" to the variable $_POST['nco'].
You then assigned the value of the variable $_POST['nco'] (still "undefine" as you stored there) to the variable $no.
You then printed the value stored in the variable $no.
It should be clear that this will always print the word "undefine".
If you want to print the value of the textbox with the name nco, fill out the form with that textbox, and in the page that process the form,
echo $_POST['nco'];
...is all you do.
You need to setup a form or something similar in order to set the $_POST variables. See this short tutorial to see how it works. If you click the submit button, your $_POST variables will be set.
what for you are using this line $_POST['nco'] = "undefine"; } ..?
and please cross check whether you are using form method as post and make sure that your text name is nco ... or else use the below code it will work.
<?php
$no = $_POST['nco'];
echo $no;
?>
<form name='na' method='post' action="<?php echo $_SERVER['PHP_SELF'];?>">
<input type='text' name='nco'>
</form>
thanks
Your action is wrong.
Change it to
action="try.php"