I have to declare an html link in my php file, i can do that in the following way
echo "<a href='razz.com'>Update</a>";
the link razz.com has a form fields(text type) in it, what i want to do is that when i open the link the form fields present in the link needs to be filled by the values that i declare in my initial php page.
the values of the form fields that needs to be filled are obtained from database.
How can i do that? Any tutorials or code snippet are appreciated.
Thank you.
Code snippet of what i am looking for:
a user fills form information which needs to go into database for storage:
the form is as follows:
<p>Title:<input type="text" name="title"/></p>
<p>Report No:<input type="text" name="rno"/></p>
<p>URL:<input type="text" name="url"/></p>
now after filling the form, the data will be stored in the database . A page with echo's successful with form information and a link to update the previously entered information will be displayed.
now again this update link contains the form structure , so instead of entering the information which was previously entered. the information already entered needs to be fetched and displayed in the form area.
You would need to use GET request parameters and the $_GET array on the page that is accepting the request to pass values to your link, but only if the http://razz.com/ index page actually accepts your URL parameters.
For instance (see the stuff and otherstuff GET keys in the URL):
echo "<a href='http://razz.com/?stuff=yes&otherstuff=yadayada'>Update</a>";
Then the razz.com index(.php):
$stuff = $_GET['stuff']; // with the link above, equal to yes
$otherstuff = $_GET['otherstuff']; // with the link above, equal to yadayada
echo "
<h2>$stuff</h2>
<p>I got $otherstuff.</p>
";
This would echo on razz.com/index(.php):
<h2>stuff</h2>
<p>I got yadayada.</p>
You could also use the $_POST array, but this would require more work and you would need to have a reason to do this (such as the data you're passing is transitory). You could do this by triggering a form, which you could also do a GET request with.
However, your question is not really that well worded. If you can provide more context and direction, that would help.
Related
I need your help!
I am new to codeigniter and I am trying to pass a fetch parameter from url, get me the value using the following
My example url is the following: http://localhost/infocargacasosfinal/index.php/creacionacta/?NroGestion=1&NroContacto=103386816
Where what I am rescuing is the value "NroContacto", I have managed to bring me the value with the following line in the controller:
'NroContacto' => $this->input->get('NroContacto');
and show it in the view with the following:
<td>
Numero Folio: <?php echo $nrocontacto; ?>
</td>
that way I have managed to show it in the user's view, but now I can't find a way to save that value in the database when the user clicks the "save" button
You have not shared much information related to your issue. I am assuming that you need two server calls to save the URL parameter into data.
First "http://localhost/infocargacasosfinal/index.php/creacionacta/?NroGestion=1&NroContacto=103386816", is used to render the page with a form to submit.
Second, When the user clicks the submit button in the form, data is stored in a database.
If this is the case, on first page, you can collect nrocontacto as you have already done in your code and add a hidden input field <input type="hidden" name="nrocontacto" value="<?php echo $nrocontacto; ?>"> inside the form in the view file.
When the user submit the form, you can grab the value using $nroContacto = $this->input->post('NroContacto', true); (this code goes to form processing controller method). Now you can use CI query builder $this->db->insert('tableName', array('field_name' => $nroContacto)); (Note: add other required fields for insert array) to save it to the Database.
Codeigniter support seo friendly urls so you can arrange your url like this
http://example.com/index.php/news/local/metro/crime_is_up
for your url it will be
http://localhost/infocargacasosfinal/index.php/creacionacta/1/103386816
Now you can fetch like this
'NroContacto' => $this->uri->segment(3, 0);
now pass the value to your view page.
You can get detailed documentation about url segments here
https://codeigniter.com/userguide3/libraries/uri.html
I hope I'm not posting a duplicate question but I've looked around (and googled as well!) and nothing has given me the answer I'm looking for.
I have a form in HTML. When the user submits the form the values get stored with mysql under their user account for the site.
The issue is, I'd like the user to be able to go back and edit the form any time they like.
I could certainly just populate the form with values from php when the users review the form, but it gets tricky when I try to populate a file input field (and the file has been saved in mysql using the blob type). Not to mention that I'd like to do this as cleanly as possible.
Ideally it would be nice if there was a convenient module for reviewing forms that have already been submitted in JQuery per se.
Can anyone offer any advice? Thanks in advance!
Edit:
Here's a good example of what I mean - in chrome if I fill out a form and redirect to the next page after hitting submit, if I hit back I come back to the form and it's still filled out with the information I entered previously! Could I invoke this behaviour whenever I want to, as opposed to only when the user hits back?
You can't pre-fil an <input type="file" . . but surely when they come back to the form, they want to see the file they've uploaded .. this is what you mean right ..
So if its a picture, you could just do: <img src="loadpic.php?id=$var" />
If it's files they've uploaded, just list the file name / date and other data.. etc in some sort of list.
Then you could still show the <input type="file"> .. but with the label, 'add more pictures' or 'add another file'. .etc
Unless someone has a better way, at the moment I'm using a combination of 2 things:
1) Utilizing the $_SESSION variable
2) Setting the "name" attribute of every input in the form to the name of the field it corresponds to in the database.
This way I can loop through all the values dynamically instead of hardcoding them all in. Some input types (like file) are exceptional and will be handled on their own. Other that I can do something like this:
To insert into mysql:
$fields = array();
$values = array();
foreach ($_POST as $field => $value) {
$fields[] = $field;
$values[] = addslashes($value);
}
$fieldString = 'Table_Name('.implode(', ', $aFields).')';
$valueString = "VALUES('".implode("', '", $aValues)."')";
mysql_query("INSERT INTO $fieldString $valueString");
Reviewing the form is somewhat similar. I am using javascript to hook into document.onload. I need to pass javascript the records from mysql so that it may populate the form. Then it's a simple matter of getting elements by their name and assigning them their values that were passed from php.
The easiest way to do it and not have to go back to the database would be to store the values in a session.
<?php $_SESSION['myvalue'] = $inputvalue; ?>
On the html form use:
<input type="text" name="myName" value="<?php echo $_SESSION['inputvalue']; ?>" />
When completed don't forget to unset the session variable:
<?php session_start(); unset($_SESSION['myvalue']); ?>
I swear i couldn't find a simple working solution for this.
On a form i have inputs that have names containing "[]" and i cant change the names of the inputs because they are part of a script.
I want to php POST the values of those inputs at the next page, after the form submit.
Example of input
<input type="text" name="CustomFields[13]" id="CustomFields_13_1" value="">
Anyone knows how to accomplish it?
I want to do it using PHP only
If the name is CustomFields[13], then you can access it at $_POST['CustomFields']['13'].
You "cannot" POST something with PHP. It's always the client that POSTs to the server. PHP is running on server side.
I recommend that you use sessions and save there the values that you need to have available in next pages.
This is how you set a session:
session_start();
$_SESSION['CustomField'] = "test";
And this is how you get it:
session_start();
echo $_SESSION['CustomField']; //Should display "test"
im using a form in php to submit some information into my database
so i used two function to do this
but how to show the result in th same page that has the form
To load the same page you have to assign the variable $_SERVER[PHP_SELF] for the form action field.
<form action='$_SERVER[PHP_SELF]?op=ban' method='post'>
then when the page get load you just check the post variable ,if it contains the appropriate data then print the result with the form.(Normally people using div tag to print the results )
It's as easy as this:
if (isset($_POST['submit']))
{
// do something with your data
}
form();
Forgive me if I am wrong. I think you have copied the code from some where and using it without understanding how forms work.
<form action='index.php?op=ban' method='post'>
The above code says to which page the values should be submitted. As you can see above the values in the form will be submitted to index.php. So the DB operations will(should) happen in index.php and the Thank you message can be shown in index.php.
If you want to show your result in the same page then you will have to submit to the page in which the form resides. But in this case you should have a logic in the page to decide whether the form was submitted or was it loaded first time.
The code snippet in your question does not tell us name of the file the code exists so we wont be able to tell you whether the result will be shown in the same page. Aslo the source code is not complete.
Post a detailed source code and we will be able to help. Hope it helps.
it should be shown on the next request.
because your app should perform an HTTP redirect after POST request.
it can be same page though
I created this layout of successive text input fields,
1- Enter data into empty fields
2- Click on button which submits to a php page that updates into database
Now the problem is that i want when i return to the main page again the empty field is replaced with data just added but there are still other empty fields to enter new data.
How can i establish that?
Thanks in advance.
You haven't given a lot of detail but here goes!
You could build your inputs like this:
<input type="text" name="age" value="<?php echo $age; ?>">
When the form first loads, it won't have values for variables like $age, so the input will appear empty. Have the form submit via POST to the same PHP file, run your validation checks, and if everything passes, insert into to your database. (Is it required that you write to the database at this point, or should it wait until the second section is filled out?)
You'll need to use some kind of conditional statement to display the second part of the form. Depending on how complex this is, or whether users will be returning later, you could:
Read the data back out of the
database to check for completeness,
and then display the second part.
Set a variable to track what stage of the form you're in, and based on that, display different sections to be completed.
If you have a way of tracking what stage of the process you're in, you could do something like this:
$formStage = 2;
function isReadOnly($formStage='')
{
if ($formStage == 2) {echo 'READONLY';}
}
and then in your HTML:
<INPUT NAME="realname" VALUE="Hi There" <?php isReadOnly($formStage)?>>