i m using isset to get the button value clicked using hidden type...below is my code
<form id="posform" name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos" value="1" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos" value="2" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos" value="3" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos" value="4" /><input type="submit" value="Update" ></div>
</form>
if (isset($_POST['updatepos'])){
$targetbtnvalue=$_POST['updatepos'];
echo $targetbtnvalue;
}
but with below code irrespective of the button clicked it always echos the last button value i.e 4.
i am trying to get it to echo the corresponding value of button clicked...plz guide
That's because it's the same form with the same field.
Try this instead:
<form name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos" value="1" /><input type="submit" value="Update" ></div>
</form>
<form name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos" value="2" /><input type="submit" value="Update" ></div>
</form>
<form name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos" value="3" /><input type="submit" value="Update" ></div>
</form>
<form name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos" value="4" /><input type="submit" value="Update" ></div>
</form>
All of your hidden input names are same i.e updatepos which shouldn't be. So you can try with different names for hidden inputs or make it array like this way updatepos[]. Then you'll get all the hidden input values on your $_POST array not just the last one.
form id="posform" name="posform" action="" method="POST" >
<div><input type="hidden" name="updatepos[]" value="1" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos[]" value="2" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos[]" value="3" /><input type="submit" value="Update" ></div>
<div><input type="hidden" name="updatepos[]" value="4" /><input type="submit" value="Update" ></div>
</form>
and then with PHP try like this,
if (isset($_POST['updatepos'])){
$targetbtnvalue=$_POST['updatepos'];
print '<pre>';
print_r($targetbtnvalue);
print '</pre>';
}
Related
I don't know why it cannot show the result. I have already put isset() Function on it, but the echo string is not come out.
<form method="POST">
<input type="radio" name="colors" value="red" checked="true">
<label id="r">Red</label>
<input type="radio" name="colors" value="green">
<label id="g">Green</label>
<input type="button" name="add" value="Add">
<input type="button" name="clear" value="Clear">
</form>
<?php
if (isset($_POST['add'])) {
if (isset($_POST['colors'])) {
$colorVal = $_POST['colors'];
echo "$colorVal";
}
}
?>
The type of your Add button must be submit also the Clear button has to be reset.
By the way that $POST['colors'] needs to be $_POST['colors'].
<form method="POST">
<input type="radio" name="colors" value="red" checked="true">
<label id="r">Red</label>
<input type="radio" name="colors" value="green">
<label id="g">Green</label>
<input type="submit" name="add" value="Add">
<input type="reset" name="clear" value="Clear">
</form>
<?php
if (isset($_POST['add'])) {
if (isset($_POST['colors'])) {
$colorVal = $_POST['colors'];
echo "$colorVal";
}
}
?>
Update: (the reset button works as expected)
<form method="POST">
<input type="radio" name="colors" value="red" checked="true">
<label id="r">Red</label>
<input type="radio" name="colors" value="green">
<label id="g">Green</label>
<input type="submit" name="add" value="Add">
<input type="reset" name="clear" value="Clear">
</form>
<form method="POST">
<input type="radio" name="colors" value="red" checked="true">
<label id="r">Red</label>
<input type="radio" name="colors" value="green">
<label id="g">Green</label>
<input type="submit" name="add" value="Add">
<input type="reset" name="clear" value="Clear">
</form>
<?php
if(isset($_POST['add']))
{
if(isset($_POST['colors']))
{
$colorVal = $_POST['colors'];
echo "$colorVal";
}
}
?>
Defines a submit button (for submitting the form) by making its type as SUBMIT
<input type="submit">
The <input type="button"> defines a clickable button (mostly used with
a JavaScript to activate a script).
Also, you have a syntax error for $_POST
I can display first name, last name and the instrument they borrowed. There is a drop down if they want to change the instrument they borrowed. Why is my code not changing the instrument that the person is changing? What could be wrong in this code? Please help. I know my code is not agreeable to sql securement, this is just an activity. database
<html>
<head>
</head>
<body>
<form method="post">
<?php
$con=mysqli_connect("localhost","root","","borrow");
$q="SELECT students.studentid,students.studfname,students.studlname,instruments.instrumentname, instruments.instrumentid from students INNER JOIN student_instrument ON students.studentid=student_instrument.id INNER JOIN instruments ON student_instrument.checkoutdate=instruments.dateacquired";
//SELECT students.studentid,instruments.instrumentid,students.studfname,students.studlname,instruments.instrumentname from
//students INNER JOIN instruments ON students.studentid=instruments.instrumentid
$t="SELECT instrumentname from instruments where dateacquired='avail'";
$r=mysqli_query($con,$q);
$e=mysqli_query($con,$t);
while($result=mysqli_fetch_array($r)){
$id=$result['instrumentid'];
$studfname=$result['studfname'];
$studlname=$result['studlname'];
$instrumentname=$result['instrumentname'];
echo $studfname." ".$studlname." ".$instrumentname." ";
echo '<input type="hidden" name="id" value='.$id.'>';
echo '<select name="inst">';
while($f=mysqli_fetch_array($e)){
$avail=$f['instrumentname'];
echo '<option>'.$avail.'</option>';
}
echo '</select>';
echo '<input type="submit" value="submit" name="submit"><br>';
}
?>
</form>
<?php
if(isset($_POST['submit'])){
$id=$_POST['id'];
$inst=$_POST['inst'];
$p="UPDATE instruments SET instrumentname='$inst' WHERE id='$id'";
$q=mysqli_query($con,$p);
if($q){
header('location:student.php');
}
}
?>
</body>
</html>
Check the generated HTML code and look for the <form> and </form> tag. Currently you have a form similar like this:
<form action="post">
<input type="hidden" name="id" value="5" />
<input type="submit" />
<input type="hidden" name="id" value="8" />
<input type="submit" />
<input type="hidden" name="id" value="9" />
<input type="submit" />
<input type="hidden" name="id" value="11" />
<input type="submit" />
<input type="hidden" name="id" value="21" />
<input type="submit" />
</form>
You generate only one form. When you submit that form you will send all the ids on the same field name "id". This means only the last id will be stored in $_POST['id']. This explains the "the last row is updated, not the first one".
To fix your problem you have to create individual forms so they will look similar to this:
<form action="post">
<input type="hidden" name="id" value="5" />
<input type="submit" />
</form>
<form action="post">
<input type="hidden" name="id" value="8" />
<input type="submit" />
</form>
<form action="post">
<input type="hidden" name="id" value="9" />
<input type="submit" />
</form>
<form action="post">
<input type="hidden" name="id" value="11" />
<input type="submit" />
</form>
<form action="post">
<input type="hidden" name="id" value="21" />
<input type="submit" />
</form>
So opening and closing the form must be happening inside the while loop.
In this
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
After I submit the form, the page refreshes and the data inside the input texts gets blank
Is it possible to keep the data even after submit?
Regards.
You can simply use ajax for submitting the form.
Or use following
<form method="POST" action=""><input type="text" class="field small-field" name="tex1" value="<?php (isset($_POST['text1]))? echo $_POST['text1] : '';" /><input type="submit" value="search" name="search"/><input type="submit" value="print" name="print"/></form>
try to echo, what ever is the variable named for your input.
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php echo $_POST['tex1'];?>" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
With php for example:
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php echo $_POST['tex1']; ?>"/>
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
If you are handling the post on the same page you could just do like this on the fields where you want the posted value to be shown:
<input type="submit" value="search" name="search" <?php if( isset( $_POST['search'] ) ){ echo "value=\"". $_POST['search'] ."\"; } ?>/>
Use this:
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php if(isset($_POST['tex1'])) echo $_POST['tex1'] ?>" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
Bascially http is statelessprotocol , hence you need to save the data some where
The simplest way in this case would be to use a conditional operator
<input type="text" class="field small-field" name="tex1" value="<?php echo (isset($_POST['search'] || $_POST['search'] )?$_POST['tex1']:''); ?>" />
I have several html textareas on my site. Each has a submit button. When a user types in one of the textareas i need to know which textarea this is. These textareas are each assigned a number taken from a mysql database. I can get the numbers out of the database, but how can I make it so that when a user types in a textarea and clicks submit the submit form knows which textarea this is. Please ask to clarify if needed. I tried my best to explain the problem. thanks.
p.s. the submit button just performs a mysql set values query. I'm using php on my site.
for example: a textarea is assigned '3.' When i submit this form i need 3 to be sent into my mysql set values query.
Use a hidden input to store a reference for each form
<input type="hidden" name="database_reference" value="<?php echo $dbId; ?>" />
Then when you submit the form $_POST['database_reference'] gives you the database id.
<input type="hidden" value="5" name="which_one" />
so for example
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="1" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="2" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="3" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="4" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="5" name="which_one" />
<input type="button" />
</form>
UPDATE:
<?php
if ($_POST){
include("db_connection.php");
mysql_query("UPDATE table SET column = '".mysql_real_escape_string($_POST['text'])."' WHERE value = ".intval($_POST['value']));
echo "done";
}
?>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="1" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="2" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="3" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="4" name="which_one" />
<input type="button" />
</form>
<form method="post" action="">
<textearea name="text"></textarea>
<input type="hidden" value="5" name="which_one" />
<input type="button" />
</form>
Assuming you have multiple <form></form> tags set up, one for each <textarea>, I would just add a hidden input field in each form. For example:
<form>
<textarea />
<input type="hidden" value="1" />
<input type="submit" />
</form>
<form>
<textarea />
<input type="hidden" value="2" />
<input type="submit" />
</form>
You can flesh it out from there, but you get the idea.
This is my HTML:
<form method="POST" action="">
<?php
$skillSubCategory = $skills->showSkills(24);
for ($i = 0; $i < count($skillSubCategory); $i++) {
?>
<input type="hidden" name="skillid" value="<?php echo $skillSubCategory[$i]['skill_id']; ?>" />
<?php echo $skillSubCategory[$i]['title']; ?>
<input type="submit" name="add" value="add" /><br />
<?php } ?>
</form>
<?php if (isset($_POST['add'])) {
echo $_POST['skillid'];
} ?>
Resulting source code:
<form method="POST" action="">
<input type="hidden" name="skillid" value="25" />
Animal Grooming
25
<input type="submit" name="add" value="add" /><br />
<input type="hidden" name="skillid" value="26" />
Dog Trainer
26
<input type="submit" name="add" value="add" /><br />
<input type="hidden" name="skillid" value="27" />
Dog Walking
27
<input type="submit" name="add" value="add" /><br />
<input type="hidden" name="skillid" value="28" />
Vet
28
<input type="submit" name="add" value="add" /><br />
<input type="hidden" name="skillid" value="29" />
Beekeeping
29
<input type="submit" name="add" value="add" /><br />
</form>
What it looks like:
I get number 29 for any button clicked. Any ideas what's wrong? Why the correct number wont show up when i click add?
You can also use the buttons themselves(without changing their values):
<input type="submit" name="skillid[25]" value="add" />
<input type="submit" name="skillid[26]" value="add" />
<input type="submit" name="skillid[27]" value="add" />
To retrieve the submitted value(its not the value in this case, its the first key of the posted array):
if(isset($_POST['skillid']) && is_array($_POST['skillid']))
{
echo key($_POST['skillid'])
}
Because when you have multiple fields with the same name attribute in a form, the last one always takes precedence (with the exception of submit buttons -- the one clicked will be the only one considered). So the last hidden input with the name skillid will always be sent to the server.
When using forms like this, you usually have to use separate forms for each button. Alternatively, change the value attribute of each button and consider that from your PHP code.
Change:
<form method="POST" action="">
to:
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
then change the condition to:
if (isset($_POST['add']) && isset($_POST['skillid'])) {
EDIT: use the <option> tag instead
<select name="skillid">
<option value="25">Animal Grooming</option>
<option value="26">Dog Trainer</option>
...
</select>
Your PHP code now will be:
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<?php
$skillSubCategory = $skills->showSkills(24);
<select name="skillid">
for ($i = 0; $i < count($skillSubCategory); $i++) { ?>
<option value="<?php echo $skillSubCategory[$i]['skill_id']; ?>"><?php echo $skillSubCategory[$i]['title']; ?></option>
<?php } ?>
</select>
<input type="submit" name="add" value="add" /><br />
</form>
if (isset($_POST['add']) && isset($_POST['skillid'])) {
echo $_POST['skillid'];
} ?>