UPDATE inside a WHILE statement - php

So, I have a page with a bunch of workorders on it. Each workorder is a row in a single table, and gets put on the page with a while() statement.
I'm trying to update each row with a simple form that I put inside the while(), and an UPDATE/WHERE statement to actually add the information to the table.
Instead of adding it to the specific row, it adds it to Every row. The only thing I can think of is that my WHERE condition is wrong, but I can't seem to figure it out. Maybe it just needs fresh eyes, or maybe I'm heading in Completely the wrong direction.
Also, any specific instructions on security, a better way to do it, etc. would be very helpful. I'm learning PHP on the fly and could use a helping hand. :)
<?php
$query = "SELECT * FROM client_information";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
$which_ad = $row['ID'];?>
<b>Name:</b> <? echo $row['billing_name']; ?> <br>
<b>Job Type:</b> <? echo $row['job_type']; ?> <br>
<b>Size:</b> <? echo $row['size']; ?> <br>
<b>Text:</b> <? echo $row['text']; ?> <br>
<b>Notes:</b> <? echo $notes; ?> <br>
<br><br>
<form action="small_update.php" method="POST">
<strong>Email Message:</strong><br>
<textarea rows="8" cols="60" name="email_message"></textarea>
<input type="submit" name="submit" value="Submit"></form>
<?
$email_message = htmlspecialchars ("{$_POST['email_message']}", ENT_QUOTES);
if (mysql_errno() != 0) {
die(mysql_error());
}
mysql_query(
"UPDATE client_information
SET email_message='$email_message'
WHERE ID='$which_ad'"
);
if (mysql_errno() != 0) {
die(mysql_error());
}
}
?>

You don't specify the id in your form:
<form action="small_update.php" method="POST">
<strong>Email Message:</strong><br>
<textarea rows="8" cols="60" name="email_message"></textarea>
<input type="hidden" name="id" value="<?php echo $which_ad; ?>">
<input type="submit" name="submit" value="Submit">
</form>
you need to also make sure you know what id was submitted:
"UPDATE client_information
SET email_message='$email_message'
WHERE ID='$_POST['id']'"
Of course, you're wide open to attacks like this as everyone else is saying. You need to look into mysqli or pdo to sanitize your input...
Ans also upon inspection you're evaluating your post data in the loop. Don't do that. Just do your evaluation before everything else is processed on the page...
<?php
if($_POST)
{
//run processing here
}
// do your fetch code here and display the forms...

Related

PHP code inside HTML value attribute

there!
I want to do a database search and display the result back to the user in a pre-populated HTML form.
I located the exact part in the code that is not working but I can't understand why PHP is not picked by the server. I'm using UwAMP.
To illustrate the problem here is my short snippet of code that I need help with:
<form id="st_reg" action="" method="POST">
Student Number:
<input type="number" name="s_num" min="1000000" max="3000000" > </br>
<input type="Submit" value="Search">
</form>
<?php
if($_SERVER['REQUEST_METHOD'] == "POST"){
if(empty($_POST['s_num'])){
$errors[] = "You forgot to enter the Student No!";
}
else{
$st_no = trim($_POST['s_num']);
}
if(empty($errors)){
//Open database connection
require('../../connect_to_database/mysql_connect.php');
//Check if the student is already in the database
$query = "SELECT * FROM student WHERE student_no = $st_no";
//Run the query
$result = mysqli_query($db_connection,$query);
if(!$result){
echo "The student does not exist!";
echo"Please <a href='index.html'>go back</a> and choose another action!";
}
elseif($result){
echo "<h2>Student Details:</h2>";
while($row = mysqli_fetch_array($result)){
echo '<form id="st_reg" action="" method="POST">
<label>Student Number:</label>
<input type="number" name = "st_number" min="1000000" max="3000000" value="<?php if(isset(\$row[\'student_no\'])) echo \$row[\'student_no\']; ?> ">
AND the PHP code inside VALUE ATTRIBUTE is not executing when it should in reality. Don't bother about GLOBAL php tags not being closed 'cause they are in the file (I'm not that dump).
Please note all this code is inside a .php file with HTML code. This is a just the processing part after the form is submitted. I saved my time by using single-quotes for echo and escaped the sigle-quotes along the way where DB access was required. I tried curly brackets around variables, echo with double-quotes escaping double-qoutes within it but none of these attempts were successful. This is strange because I can perfectly echo $row['student_no'] outside of this context and is running fine.
I also looked at similar questions on this website. They were close but none of them had nearly to this context. I am open to any suggestions and better than that solutions.
echo '<form id="st_reg" action="" method="POST">
<label>Student Number:</label>
<input type="number" name = "st_number" min="1000000" max="3000000" value="<?php if(isset(\$row[\'student_no\'])) echo \$row[\'student_no\']; ?> ">
should look like this:
echo '<form id="st_reg" action="" method="POST">
<label>Student Number:</label>
<input type="number" name = "st_number" min="1000000" max="3000000" value="' . (isset($row['student_no']) ? $row['student_no'] : '') . '">
CONTINUATION OF STRING...
The following will do what you want.
value="<?= (isset($row["student_no"]) ? $row["student_no"] : "") ?>"
You don't need to worry about all of the escaping when you're inside the PHP chunk already.

PHP Delete record from database MySql

I need to delete a record, in this case a categories from my forum, from the database based on its id.
<?php
if(isset($_SESSION['signed_in']) && $_SESSION['user_level'] == 1)
{
?>
<td>
<form method="post">
<input type="hidden" value="<?= ['cat_id']; ?>">
<input type="submit" name="submit" value="Remover" />
</form>
<?php
if(isset($_POST['submit']))
{
mysql_query("DELETE FROM categories where cat_id = 'cat_id'");
}
?>
</td>
<?php
}
?>
i cant get a "good" way to do it... :(
EDIT: This is for a programming lesson not a real forum!!
Your HTML Input Field needs a name so it can be identified by your PHP.
Then, in your Code Block where you attempt to delete the category, you need to acces the category id using the $_POST array.
Another thig you want to do is read up onj the dangers of SQL injections.
If you're just playing around with PHP and MySQL at the moment: Go Ahead. But if you actually want to develop, maybe you should read up on a few other things as well, even if it seems like overkill at first: PHP The Right Way.
Nontheless, try this:
<?php
if(isset($_SESSION['signed_in']) && $_SESSION['user_level'] == 1)
{
?>
<td>
<form method="post">
<input type="hidden" name="hid_catid" id="hid_catid" value="<?php echo $cat_id; ?>">
<input type="submit" name="submit" value="Remover" />
</form>
<?php
if(isset($_POST['submit']))
{
$query = "DELETE FROM categories where cat_id = '".(int)$_POST['hid_catid']."'";
mysql_query($query);
}
?>
</td>
<?php
}
?>
--> hidden field should have name and id to use
--
Thanks
Your hidden input field needs a name to be accessable after the post. Also I am not sure if ['cat_id'] is the correcty way to reference this variable. Where does it come from?
<form method="post">
<input type="hidden" name="cat_id" value="<?= $cat_id ?>">
<input type="submit" name="submit" value="Remover" />
</form>
Then your query has to look like this to correctly grab the id from the post.
mysql_query("DELETE FROM categories where cat_id = " . mysql_real_escape_string($_POST['cat_id']));

echoing data from mysql_fetch_array

I'm trying to display data from my database table selected from a 'date'.
The query executes, but when I echo I don't get any result. Could you please help me with this?
<?php include 'includes/connection.php'; ?>
<html>
<head>
<title> </title>
</head>
<body>
<?php
if(isset($_POST['submitted'])){
$sql = "SELECT * FROM dagtaken WHERE datum = $_POST[datum]";
$result = mysql_query($sql) or die (mysql_error());
while ($row = mysql_fetch_array($result)){
echo $row['aantal'];
}
}else{
?>
<form action='' method='POST'>
<p><input type="date" name="datum"></p>
<p><input type='submit' value='Dagtaak toevoegen' />
<input type='hidden' value='1' name='submitted' /></p>
</form>
<?php } ?>
</body>
</html>
The query shouldn't execute, since dates are very obviously strings and require quotes. That said...
Try this:
mysql_query("SLEECT * FROM `dagtaken` WHERE `datum`='".mysql_real_escape_string($_POST['datum'])."'");
Now on to the actual problem, you are checking if(isset($_POST['submitted'])), but nowhere do I see <input name="submitted" in your source (EDIT Never mind, it has a stupid amount of whitespace pushing it off the edge). Try if(isset($_POST['datum'])), since that's the variable you actually use.
You haven't named your submit button, so your PHP code never kicks in. Don't check for form fields when all you really need is to check if a POST has occured.
Quick fix for you code:
<input type="submit" name="submitted" value="Dagtaak toevoegen" />
^^^^^^^^^^^^^^^^^
Better fix:
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
your code here ...
}
First, Escape your data. SQL injection is now very easy
Second, do you have data in your database?
Try print_r($row) instead of echo $row...
$_POST is with quotes...=> $_POST["datum"]
Last addition, is your date the same as your input?

populating text fields from the sql using dropdown list Jquery

Hello there first time doing this, Basically I am rather confused on how to Re-populate text boxes from the database.
My current issue is that basically I have two tables in my database 'USER' and 'STATISTICS'.
Currently what is working is that my code is looking up the values of 'User_ID' in the 'USER' table and populating the values in the drop down list.
What I want from there is for the text fields to populate corresponding to those values from the database looking up the 'User_ID' E.G 'goal_scored' , 'assist', 'clean_sheets' and etc.
I am pretty baffled I have looked up on various different questions but cannot find what im looking for.
<?php
$link = mysql_connect("localhost","root","");
mysql_select_db("f_club",$link);
$sql = "SELECT * FROM user ";
$aResult = mysql_query($sql);
?>
<html>
<body>
<title>forms</title>
<link rel="stylesheet" type="text/css" href="css/global.css" />
</head>
<body>
<div id="container">
<form action="update.php" method="post">
<h1>Enter User Details</h1>
<h2>
<p> <label for="User_ID"> User ID: </label> <select id="User_ID" id="User_ID" name="User_ID" >
<br> <option value="">Select</option></br>
<?php
$sid1 = $_REQUEST['User_ID'];
while($rows=mysql_fetch_array($aResult,MYSQL_ASSOC))
{
$User_ID = $rows['User_ID'];
if($sid1 == $id)
{
$chkselect = 'selected';
}
else
{
$chkselect ='';
}
?>
<option value="<?php echo $id;?>"<?php echo $chkselect;?>>
<?php echo $User_ID;?></option>
<?php }
?>
I had to put this in because everytime I have text field under the User_ID it goes next to it and cuts it off :S
<p><label for="null"> null: </label><input type="text" name="null" /></p>
<p><label for="goal_scored">Goal Scored: </label><input type="text" name="Goal_Scored" /></p>
<p><label for="assist">assist: </label><input type="text" name="assist" /></p>
<p><label for="clean_sheets">clean sheets: </label><input type="text" name="clean_sheets" /></p>
<p><label for="yellow_card">yellow card: </label><input type="text" name="yellow_card" /></p>
<p><label for="red_card">red card: </label><input type="text" name="red_card" /></p>
<p><input type="submit" name="submit" value="Update" /></p></h2>
</form>
</div>
</body>
</html>
If anyone can help with understanding how to get to the next stage would be much appreciated thanks x
Rather than spending time on something complicated like AJAX, I'd recommend going the simple route of pages with queries, such as user.php?id=1.
Craft a user.php file (like yours) and if id is set (if isset($_GET['id'])) select that user from the database (after having sanitised your input, of course) with select * from users where id = $id (I of course assume you have an id for each user).
You can still have the <select>, but remember to close it with </select>. You might end up with something like this:
<form method="get">
<label for="user">Select user:</label>
<select name="id" id="user">
<option value="1">User 1</option>
...
</select>
<submit name="submit" value="Select user" />
</form>
This will send ?id=<id> to the current page and you can then fill in your form. If you further want to edit that data, create a new form with the data filled in with code like <input type="text" name="goal_scored" value="<?php echo $result['goal_scored']; ?>" /> then make sure the method="post" and listen on isset($_POST['submit']) and update your database.
An example:
<?php
// init
// Use mysqli_ instead, mysql_ is deprecated
$result = mysqli_query($link, "SELECT id, name FROM users");
// Create our select
while ( $row = mysqli_fetch_array($link, $result, MYSQL_ASSOC) ) {?>
<option value="<?php echo $result['id']; ?>"><?php echo $result['name'] ?></option>
<?php}
// More code ommitted
if (isset($_GET['id'])) {
$id = sanitise($_GET['id']); // I recommend creating a function for this,
// but if only you are going to use it, maybe
// don't bother.
$result = mysqli_query($link, "SELECT * FROM users WHERE id = $id");
// now create our form.
if (isset($_POST['submit'])) {
// data to be updated
$data = sanitise($_POST['data']);
// ...
mysqli_query($link, "UPDATE users SET data = $data, ... WHERE id = $id");
// To avoid the 'refresh to send data thing', you might want to do a
// location header trick
header('Location: user.php?id='.$id);
}
}
Remember, this is just an example of the idea I'm talking about, lots of code have been omitted. I don't usually like writing actually HTML outside <?php ?> tags, but it can work, I guess. Especially for smaller things.

passing an array through the function

My website is featuring online classified advertisements, programmed by PHP and MySQL. The following code let the administrator delete multiple records using the checkbox tool.
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >
<table>
<td><? echo $rows['CountryName']; ?></td>
<td><input name="delete_items[]" type="checkbox" value="<?php echo $rows['id']; ?>" /></td>
</table>
<input type="submit" name="deleteSelected" value="Submit" >
</form>
<?php
if(isset($_POST['deleteSelected'])) {
$delete_items = join(', ', $_POST["delete_items"]);
$query = "DELETE FROM $table_name WHERE id IN ($deleted_items)";
$result = mysql_query($query);
header("Location: admin.php"); }
?>
When I press the submit button without checking boxes (all boxes are unchecked), I receive the following error message (p.s. the script is working well without any error message, if any Checkbox being checked):
Warning: join() [function.join]: Invalid arguments passed in C:\xampp\htdocs\admin_listing.php on line 87
I’ve tried the implode function instead of using join, but still I'm getting an error message.
Maybe I do not passing an array through the function correctly, but I cannot find a solution for the above.
Any advise would be appreciated.
It looks like you are displaying all the records from your database into a single input in the form.
The code will probably work well with the implode() as you tried, but you will need to use a loop in the displaying of the form to generate it properly with the options.
Something like this:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >
<table>
<?php
while($row=$databaseResult) //however you are getting the data out of the database.
{
echo "<tr><td>".$rows['CountryName']."</td><td><input name='delete_items[]' type='checkbox' value=".$rows['id']."/></td></tr>";
}
?>
</table>
<input type="submit" name="deleteSelected" value="Submit" >
</form>
Thank you all for trying to help, I found a simple solution by adding one code line, as follows:
<?php
if(isset($_POST['deleteSelected'])) {
if(isset($_POST["delete_items"][0])) {
$delete_items = join(', ', $_POST["delete_items"]);
$query = "DELETE FROM $table_name WHERE id IN ($delete_items)";
$result = mysql_query($query);
header("Location: admin.php");
}
}
?>
Hope it can help someone else...

Categories