I have a created an HTML form where users sign up and input there data into an SQL database. I have then retrieved that data in a webpage where they can view their profile. I have created a page where users can edit there profile by creating a form which updates the value in the SQL database for there user id.
I would like the form to use the current value of the SQL cell as the default for that user to make alterations easier. Example: currently user 7 has their city set as New York, when they visits the edit info page, the city field in the form already hase New York as the default value.
I have no problem getting the SQL info and assigning it to a variable, I just don't understand how to set it as the default value. I am aware of how you set default values for input fields though.
My code:
<?php
$id = $_SESSION["user_id"];
// Create a query for the database
$query = "SELECT full_name FROM users WHERE id = $id LIMIT 1";
// Get a response from the database by sending the connection
// and the query
$response = #mysqli_query($dbc, $query);
// If the query executed properly proceed
if($response){
while($row = mysqli_fetch_array($response)){
echo $row['full_name'];
echo mysqli_error();
}
}
?>
<input type="text" name="aboutme" defualt="<?php echo $row['aboutme'] ?>" >
There's no default value for html input.
Input can has value, using attribute value:
<input type="text" name="some_name" value="Some value" />
In your case it's
<input type="text" name="aboutme" value="<?php echo $row['aboutme']?> />
Input can also has placeholder - some value that is present in an input, but erased when user starts to edit input's content:
<input type="text" name="aboutme" value="<?php echo $row['aboutme']?> placeholder="some value" />
How about
<?php
$id = $_SESSION["user_id"];
// Create a query for the database
$query = "SELECT full_name FROM users WHERE id = $id LIMIT 1";
// Get a response from the database by sending the connection
// and the query
$response = #mysqli_query($dbc, $query);
// If the query executed properly proceed
if($response){
while($row = mysqli_fetch_array($response)){
echo $row['full_name'];
?>
<input type="text" name="aboutme" value="<?php echo $row['aboutme'] ?>" >
<?php
echo mysqli_error();
}
}
?>
And here is a good example http://www.w3schools.com/php/showphpfile.asp?filename=demo_db_select_pdo
Neither of the answers worked and upon further research and trial and error I created a solution.
I changed the value that was store in the array to just be a normal php variable:
$aboutme = $row['aboutme'];
I then called that variable using the following code:
<input type="text" name="aboutme" value="<?php echo htmlspecialchars($aboutme); ?>" >
Thanks for your help.
I hope you find my answer useful.
Why don't you try using it as a place holder? This will provide editable text.
<input type="text" name="aboutme" placeholder="<?php echo $row['aboutme'];" />
Related
I am trying to increment project number based on the last entry. The the primary key PROJECTNOID auto-increments but is not the same format as the project number (Ex: PROJECTNOID = 1 and Project Number = 19000). I don't want this to be a dropdown box even though some of my code shows the opposite.
<?php
connect = mysqli_connect("**", "**", "**", "**");
$query4 = "SELECT PROJECTNOID, ProjectNumber FROM tblProjects ORDER BY
PROJECTNOID";
$result4 = mysqli_query($connect,$query4);
$options4 = "";
while($row4 = mysqli_fetch_row($result4);){
$options4 = $options4."<input value=$row4[0]$row4[1]</input>";
}
?>
Here is the html textbox:
<label for="txtfield">Project Number</label>
<!--<input type="text" id="reqtxtfield" name="projectnumber"
value="<?php ?>" readonly/>-->
<?php echo $options4;?>
But it would look like how you had it but instead of '1' inside the
box it would display '19000' and there would be nothing outside of the
box other than the label "Project Number". As far as i'm aware you can
assign a value to the text box, regardless of whatever the input is. I
would like it to display the value from one field name but actually
contain the value from a different field name. Both are in the same
table of course.
OK - gotcha. Unfortunately, you cannot do that. A textbox can only have one value and the user is always free to change that value, even if you make it read-only. You can test that out by using the developer toolbar in your browser. Probably a good time to mention that all user input should be considered dangerous and you should never trust it. Once they have submitted the form you need to verify it.
What I would recommend in your case is to use a hidden <input> which contains the value you actually want to submit; projectnoid. You can then display the Project Number in any manner you choose.
<form>
<h1>Project Number: 19000</h1>
<input type="hidden" name="projectnoid" value="1">
<input type="submit" name="submit">
</form>
To generate this, you would:
<?php
while($row4 = mysqli_fetch_row($result4)){
$projectnoid = $row[0];
$projectNumber = $row[1];
echo '<h1>' . $projectNumber . '</h1>';
echo '<input type="hidden" name="projectnoid" value="'. $projectnoid .'">
}
PHP:
<?php
$query_5 = "SELECT MAX(ProjectNumber) FROM tblProjects;";
$result_5 = mysqli_query($conn, $query_5);
$row_5 = mysqli_fetch_array($result_5);
$nextproject=$row_5['MAX(ProjectNumber)']+1;
?>
HTML:
<html>
<form class="myform" action="<?php echo htmlspecialchars($_SERVER[" PHP_SELF "]);?>" method="post">
<label for="txtfield">Project Number</label>
<input type="text" id="reqtxtfield" name="projectnumber" value="<?php echo $nextproject ?>" readonly/><br>
After running successful insert query:
echo "<meta http-equiv='refresh' content='0'>"; //REFRESH PAGE TO UPDATE PROJECT NUMBER
So I have searched for how to insert items from a MySQL database into a form. From what I understand, for each field, I need to add value="<?php echo htmlentities($Variable); ?>" is there a way to insert the fields all at once? Like in a PHP script? Each field already has a name identifier, that's how the insert.php insert the values into the database, like this:
//SQL Query to insert data
$sql="INSERT INTO case_info (Volunteer,CaseShortTitle,CaseNumber,HearingDate...)
VALUES ('$_POST[Volunteer]','$_POST[CaseShortTitle]','$_POST[CaseNumber]','$_POST[HearingDate]'...)
and the form fields look like this:
<div><input type="text" id="p1f5c" name="CaseShortTitle" class="fieldcontent"><div class="fielderror"></div></div>
Then the Submit button goes to this insert.php file
Is there a way to use GET to populate the fields in this manner?
Thank you!
It's generally bad practice to input data into your database directly from the $_POST variables, you should consider sanitizing them first.
If I understand what you are asking, you can do something like this;
<form action="" id="myForm" title="myForm">
<?php
$query = "SELECT * FROM case_info LIMIT 1";
$result = myqsl_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
?>
<input type="text" id="volunteer" value="<?php echo $row['Volunteer']; ?>"
<input type="text" id="CaseShortTitle" value="<?php echo $row['CaseShortTitle']; ?>"
<?php
} //Close while{} loop
?>
</form>
The way I've structured my form data is by creating them in a while loop, but each time they are created the form will take a unique id.
So my question is, how do I access them individually and update specified data to a MYSQL server.
I've attempted to do it in the code at the end of the script, but I'm not sure how to access the forms individually
<?php
include 'user_data.php';
include 'core.inc.php';
$query = mysql_query("SELECT `post_text` FROM `posts`,`sub_posts` WHERE sub_posts.post_id = posts.id AND sub_posts.user_id='$user_id'");
while($row = mysql_fetch_array($query)){
?><?php echo $row[post_text].'<br>'?>
<form action="<?php $curent_file ?>" method="POST">
<textarea name="answer_field" > </textarea><br />
<input type="submit" value="Submit Answer">
<input type="hidden" name="post_id" value="<?php echo $row['post_id']; ?>" />
</form>
<?php
}//While Loop
if (isset($_POST['answer_field']) && !empty($_POST['answer_field'])){
$answer = mysql_real_escape_string($_POST['answer_field']);
$id = intval($_POST ['post_id']);
$query = "UPDATE `sub_posts` SET `sub_answer`='$answer' WHERE `post_id`='$id'";
}
?>
Only a single form gets posted when clicking the "submit" field. The form name does not get submitted by itself. Instead, you would place the post ID to which the form corresponds as a hidden field:
<input type="hidden" name="post_id" value="<?php echo $row['post_id']; ?>" />
And then later:
$answer = mysql_real_escape_string ($_POST ['answer']);
$id = intval ($_POST ['post_id']);
$query = "UPDATE `sub_posts` SET `sub_answer`='{$answer}' WHERE `post_id`={$id}";
Note that you definitely need to escape the answer before putting it in the query and make sure that the ID is a number. Otherwise, you're opening up your code to SQL injection attacks.
I have a simple Form along side a PHP update query that simply isn't working! I know the PHP is working on the page as there are several validation checks that need to be passed before hand which are working perfectly. The form its self is inside the Colorbox Popup tool.
My HTML Form Code is:
<div id="stylized" class="myform">
<form action="#" method="post">
<input type="hidden" name="user_id" value="<?php echo $user_id; ?>" />
<label>First Name:<span class="small">Enter your forename</span></label>
<input id="first_name" type="text" name="first_name" maxlength="50" placeholder="e.g. Joe" required autofocus/>
<div class="spacer"></div>
<input type="submit" id="update" name="update" value="Continue to Step 2!">
</form>
</div>
With the PHP Code as follows (this is above the HTML code on the page):
<?php
if($_POST['update']){
$user_i = $_POST['user_id'];
$f_name = $_POST['first_name'];
$first_name = ucfirst($f_name);
mysql_query("UPDATE user SET first_name = '$first_name' WHERE user_id = '$user_i'") or die(mysql_error());
} ?>
The actual submit appears to be working, with the Popup refreshing afterwards, but the database does not update! I have triple checked the syntax and the database fields. 'user' and 'first_name' and 'user_id' is correct.
Update: Because the popup box refreshes, I cannot view the error's from the 'or die(mysql_error()) unfortunately, other wise i might have been one step closer.
Any help would be hugely appreciated.
Many thanks in advance.
When you say pop-up box, I assume you are using ajax to communicate from the form to the server, which as you stated is difficult to view submitted data. If this is the case try:
error_log(serialize($_POST));
This will force an entry in your error log with the $_POST data in serialized format, so you can check the values you are submitting are populated correctly.
You will also want to sanitize the variables you are adding to the SQL:
$sql = "UPDATE user SET first_name = " . mysql_real_escape_string($first_name) . " WHERE user_id = " . mysql_real_escape_string($user_i) . " LIMIT 1";
mysql_query($sql);
I would:
print_r($_POST); to view the POST data.
Generate the SQL from a string so it can be printed for debugging purposes, like so:
$sql = "UPDATE user SET first_name = '$first_name' WHERE user_id = '$user_i'";
echo $sql;
mysql_query($sql) or die(mysql_error());
One of these techniques will likely tell you why the PHP-generated SQL doesn't update your database record.
you set your user_id field by echo $user_id; but your variable name is set to $user_i = $_POST['user_id'];
therefore your user id field is not set and your Mysql command will fail.
Well i heard that Value attribute is just a default attribute, so i used the value attribute to fetch the data from a previous file and fill form area.
But it is only using value of value attribute() to store in DB. For Example if the customer changes value passed in the text box(passed by value attribute) and submit the information the changed value is not being stored in DB rather value stored in value attribute is getting saved.
I am passing values to a html form, from a previous PHP page like this..
<form action="parentid1.php" method="post">
<label>id:</label>
<input type="text" name="id" value="<?php echo htmlspecialchars($_GET['id']) ?>" size="50"/> <br/>
<label>Name:</label>
<input type="text" name="name" value="<?php echo htmlspecialchars($_GET['name']) ?>" size="50"/><br/>
<input type="submit" value="submit" /><br/>
</form>
In my next file i am trying yo update the values declared like this:
$value = mysql_real_escape_string($_POST['id']);
$value1 = mysql_real_escape_string($_POST['name']);
$sql = "UPDATE table SET name='$value1'WHERE doi ='$value'";
Here is the case:
A form structure of text type will be formed with some value already in it.
Now customer wants to change the data in form box changes it and submits it but the changed is not being saved in the DB rather the previous value is only saved.
What am i doing wrong?
It seems that you mix GET and POST data. In html you use GEt but in php POST.
If you don't know about GET or POST try to use $_REQUEST.
$sql = "UPDATE `table` SET `name`='".$value1."' WHERE `doi`='".$value."'";
$result = mysql_query($sql) or die(mysql_error());
and show your mysql error.