mysqli_query of update statement isn't working in PHP - php

I have written code with an update statement but it doesn't affect the database. I have tried with and without single quotes but it still doesn't update the database. I have by default value of Details="Order Placed" in the table but when I update the value with the update query it doesn't work.
<?php
include("connection1.php")
?>
<form method="POST">
Enter Orderid to be updated<br>
<input type="text" name="id">
<br>
<input type="radio" id="m" name="info" value="placed">
<label for="m">Order Placed</label><br>
<input type="radio" id="f" name="info" value="shipped">
<label for="f">Order Shipped</label><br>
<input type="radio" id="o" name="info" value="Delivered">
<label for="o">Order Delivered</label>
<input type="submit" name="submit" value="submit">
</form>
<?php
if(isset($_POST['submit']))
{
$i=$_POST['id'];
$info1=$_POST['info'];
$query=("SELECT * FROM shipments WHERE OrderId='$i'");
$data=mysqli_query($conn, $query);
$total=mysqli_num_rows($data);
if($total==1)
{
$query1=mysqli_query("UPDATE `shipments` SET `Details`='$info1
WHERE`OrderId`='$i'");
if($result==mysqli_query($conn, $query1))
{
echo "ORDER UPDATED";
}
else
{
echo "ORDER NOT UPDATED";
}
}
}
?>

Just remove unwanted mysqli_query in Update statement $query1...
Change
$query1=mysqli_query("UPDATE `shipments` SET `Details`='$info1
WHERE`OrderId`='$i'");
To
$query1= "UPDATE shipments SET Details ='$info1' WHERE OrderId='$i'";

Related

How to display information of a selected user from a database for editing

I Am trying to edit user details from a database. I have queried the database and stored the info in $row, but each time I try to echo the details I get the following error: Trying to access array offset on the value of type null in
C:\xampp\htdocs\merchant\admin\edituserhis.php
I seem not to know what I am doing wrong in the SQL statement, I have checked the Variable $id and its working well.
<div class="quotes">
<?php
$con = mysqli_connect("localhost","root","","merchant_db");
$id = $_REQUEST['username'];
$query = "SELECT * from user_trans where userid='".$id."'";
$result = mysqli_query($con, $query) or die ( mysqli_error());
$row = mysqli_fetch_assoc($result);
?>
<?php echo $row['userid']; ?>;
<form name="form" method="post" action="userprofile.php">
<input type="hidden" name="new" value="1" />
<input name="id" type="hidden" value="<?php echo $row['userid'];?>" />
<p><input type="text" name="balance" placeholder="Enter Amount" required value="<?php echo $row['description'];?>" /></p>
<p><input name="submit" type="submit" value="Credit" /></p>
</form>
</div>

How to store and get two values in PHP?

This is process_upcategory.php
I want to update the category name or the category id with another category name/id by its category id or by its category name.
I'm new to php
<?php
require('includes/config.php');
if(!empty($_POST))
{
$msg=array();
if(empty($_POST['cat']))
{
$msg[]="Please full fill all requirement";
}
if(!empty($msg))
{
echo '<b>Error:-</b><br>';
foreach($msg as $k)
{
echo '<li>'.$k;
}
}
else
{
$cat_nm=$_POST['cat[0]'];
$cat_id=$_POST['cat[1]'];
$query= "UPDATE `category` SET cat_nm='$cat_nm' WHERE cat_id='$cat_id'";
mysqli_query($conn,$query) or die("can't Execute...");
mysql_close($link);
header("location:category.php");
}
}
else
{
header("location:index.php");
}
?>
Now this is category.php, just a snippet of code. Not whole code
<form action='process_upcategory.php' method='POST'>
<b style="color:darkgreen">UPDATE CATEGORY </b> <br>
<b style="color:darkgreen">Old Category</b>
<br>
<select name="cat[]" multiple>
<?php
$query="select * from category ";
$res=mysqli_query($conn,$query);
while($row=mysqli_fetch_assoc($res))
{
echo "<option>".$row['cat_nm'];
echo "<option>".$row['cat_id'];
}
?>
</select>
<br>
<b style="color:darkgreen">New Category</b><br>
<input type='text' name='cat[0]'></input><br>
<input type='text' name='cat[1]'></input>
<input type='submit' value=' UPDATE '>
</form>
I want to update the category name with another category name by its category id or by its category name. I get undefined index cat[0] and cat[1]
When you end an input name with [] it wil be converted to an array by php. The correct way to get the values in this case would be something like this:
$cat=$_POST['cat'];
$cat_nm=$cat[0];
$cat_id=$cat[1];
I combined the two routines into one script.
I added 'sub' to the form to distinguish from when the form was submitted or not.
I used list() in the query result loop.Used mysqli_fetch_array($result, MYSQLI_NUM) rather than mysqli_fetch_assoc($res)
used foreach() to loop through the $_POST['cat']
Added 'value' to the <option value=""> to hold the id
Eliminated the switching from HTML mode to PHP mode by using HEREDOC.
<?php
if (intval($_POST['sub']) == 1){
$newcat = $_POST['new'];
foreach($_POST['cat'] as $key=>$value){
if(strlen($newcat[$key]) > 0){
mysqli_query($conn,"UPDATE `category` SET `cat_nm`='$newcat[$key]' WHERE `cat_id`='$value'");
}
}
}
echo <<<EOT
<html><head><style>h4,h3{color:darkgreen;margin:.2em;}</style></head><body>
<form action="#" method='POST'>
<h3>UPDATE CATEGORY</h3>
<h4>Old Category</h3>
<select name="cat[]" multiple>
EOT;
$sql="SELECT `cat_nm`, `cat_id` FROM `category` ";
$result=mysqli_query($conn,$sql);
while(list($cat_nm,$cat_id) = mysqli_fetch_array($result, MYSQLI_NUM)){
echo " <option value=\"$cat_id\">$cat_nm</option>\n";
}
echo <<<EOT
</select>
<h3>New Category</h3>
<input type="text" name="new[0]" /><br/>
<input type="text" name="new[1]" /><br/>
<input type="hidden" name="sub" value="1" /><br/>
<input type="submit" value=" UPDATE />
</form>
</body></html>
EOT;
?>

Update multiple SQL records using one PHP/MySQLi query

Is it possible to update multiple records in one MySQLi query?
There are 4 records to be updated (1 for each element level) when the submit button is clicked.
The results are posted to a separate PHP page which runs the query and returns the user back to the edit page. elementid is 1,2,3,4 and corresponds with Earth, wind, fire, water. These never change (hence readonly or hidden)
<form id="edituser" name="edituser" method="post" action="applylevelchanges.php">
<fieldset>
<legend>Edit Element Level</legend>
<?php
while($userdetails->fetch())
{?>
<input name="clientid" id="clientid" type="text" size="8" value="<?php echo $clientid; ?>" hidden />
<input name="elementid" id="elementid" type="text" size="8" value="<?php echo $elementid;?>" hidden />
<input name="elemname" id="elemname" type="text" size="15" value="<?php echo $elemname; ?>" readonly />
<input name="elemlevel" id="elemlevel" type="text" size="8" required value="<?php echo $elemlevel; ?>" /></br>
</br>
<?php }?>
</fieldset>
<button type="submit">Edit Student Levels</button>
</form>
And the code to apply the changes
<?php
if (isset($_POST['clientid']) && isset($_POST['elementid']) && isset($_POST['elemname']) && isset($_POST['elemlevel'])) {
$db = createConnection();
$clientid = $_POST['clientid'];
$elementid = $_POST['elementid'];
$elemname = $_POST['elemname'];
$elemlevel = $_POST['elemlevel'];
$updatesql = "update stuelement set elemlevel=? where clientid=? and elementid=?";
$doupdate = $db->prepare($updatesql);
$doupdate->bind_param("iii", $elemlevel, $clientid, $elementid);
$doupdate->execute();
$doupdate->close();
$db->close();
header("location: edituserlevel.php");
exit;
} else {
echo "<p>Some parameters are missing, cannot update database</p>";
}

html radio button with an "other" selection that has a text box

When my insert runs it will write the text in the text box but if I choose a radio button selection other than "other" nothing is inserted into the table.
If I remove the "other"code the radio button selected is written to the table.
here is the code
<input type="radio" name="job" value="PHP Programmer">PHP Programmer
<input type="radio" name="job" value="SQL Programmer">SQL Programmer<br>
<input type="radio" name="job">Other <input type="text" name="job" >
<h3>* If yes, check which ISO standard(s) you are accredited?</h3>
<input type="radio" name="iso_standard" value="17020">17020
<input type="radio" name="iso_standard" value="17025">17025
<br />
<input action="submit" type="submit" name="submit" value="Submit">
</form>
<?php
if(isset($_POST['job']) && isset($_POST['iso_standard']))
{
$job = mysqli_real_escape_string($db, $_POST['job']);
$iso_standard =mysqli_real_escape_string($db, $_POST['iso_standard']);
$sql="insert into tbl_test_insert(iso_cert, iso_standard)
values ('$job', '$iso_standard')";
if(!mysqli_query($db, $sql))
{
die('Error: ' .mysqli_error($db));
}
echo "1 record added";
}
else
{
echo "You didn't choose all the options!
}
?>
You are overwriting the value of job with the text input field. It does not mather if it has any value or not.
Please rename your <input type="text" name="job" > to for example otherJob and you should be fine
edit:
change <input type="radio" name="job">Other <input type="text" name="job" >
to <input type="radio" name="job" value="other">Other <input type="text" name="otherJob">
change $job = mysqli_real_escape_string($db, $_POST['job']);
to
$job = mysqli_real_escape_string($db, $_POST['job']);
$otherJob = mysqli_real_escape_string($db, $_POST['otherJob']);
if ($job == 'other') {
$jobField = $otherJob;
} else {
$jobField = $job;
}
and change $sql="insert into tbl_test_insert(iso_cert, iso_standard)
values ('$job', '$iso_standard')";
to $sql="insert into tbl_test_insert(iso_cert, iso_standard)
values ('$jobField', '$iso_standard')";

Table not updating after mysql query

I have an administrator.php which displays 300 records from a table called 'player'. Next to each record, there is an edit option which redirects you to edit.php and the 15 columns of that record (including the primary key - playerid) is displayed inside text boxes. Line of code below:
<a href='edit.php?playerid=".$query2['playerid']."'>Edit</a>
On edit.php you are able to change data of these columns. Upon submit, an update query is sent to update the table but unfortunately, it's not working. My error message continues to display ("testing for error..."); not sure why.
//Setups up the database connection
$link = mysql_connect("localhost", "root", "");
mysql_select_db("fantasymock", $link);
if(isset($_GET['playerid'])) {
$playerid = $_GET['playerid'];
//Query to display results in input box
$query1 = mysql_query("SELECT * from player WHERE playerid = '$playerid'");
$query2 = mysql_fetch_array($query1);
}
if(isset($_POST['submit'])) {
$playerid = $_POST['playerid'];
$preranking = $_POST['preranking'];
$playerlast = $_POST['playerlast'];
$playerfirst = $_POST['playerfirst'];
$position = $_POST['position'];
$battingavg = $_POST['battingavg'];
$run = $_POST['run'];
$homerun = $_POST['homerun'];
$rbi = $_POST['rbi'];
$sb = $_POST['sb'];
$win = $_POST['win'];
$save = $_POST['save'];
$strikeout = $_POST['strikeout'];
$era = $_POST['era'];
$whip = $_POST['whip'];
//Query to update dB
$query3 = mysql_query("UPDATE player SET playerid='$playerid', preranking='$preranking', playerlast='$playerlast', playerfirst='$playerfirst', position='$position', battingavg='$battingavg', run='$run', homerun='$homerun', rbi='$rbi', sb='$sb', win='$win', save='$save', strikeout='$strikeout', era='$era', whip='$whip' WHERE playerid='$playerid'");
header("Location: administrator.php");
} else {
echo "Testing For Error....";
}
?>
<form action="" method="POST">
Player ID:<input type="text" name="playerid" value="<?php echo $query2['playerid'];?>"/> <br/>
Preranking:<input type="text" name="preranking" value="<?php echo $query2['preranking'];?>"/> <br/>
Last Name:<input type="text" name="playerlast" value="<?php echo $query2['playerlast'];?>"/> <br/>
First Name:<input type="text" name="playerfirst" value="<?php echo $query2['playerfirst'];?>"/> <br/>
Position:<input type="text" name="position" value="<?php echo $query2['position'];?>"/> <br/>
Batting Avg:<input type="text" name="battingavg" value="<?php echo $query2['battingavg'];?>"/> <br/>
Runs:<input type="text" name="run" value="<?php echo $query2['run'];?>"/> <br/>
Homeruns:<input type="text" name="homerun" value="<?php echo $query2['homerun'];?>"/> <br/>
Rbi:<input type="text" name="rbi" value="<?php echo $query2['rbi'];?>"/> <br/>
Sb:<input type="text" name="sb" value="<?php echo $query2['sb'];?>"/> <br/>
Wins:<input type="text" name="win" value="<?php echo $query2['win'];?>"/> <br/>
Saves:<input type="text" name="save" value="<?php echo $query2['save'];?>"/> <br/>
Strikeouts:<input type="text" name="strikeout" value="<?php echo $query2['strikeout'];?>"/> <br/>
Era:<input type="text" name="era" value="<?php echo $query2['era'];?>"/> <br/>
Whip:<input type="text" name="whip" value="<?php echo $query2['whip'];?>"/> <br/>
<br>
<input type="submit" name="submit" value="submit">
</form>
FYI: Every column in the table and tablename is spelled correctly, I've triple checked before posting. And I'm aware of MySQL injection. Can someone see a problem? Thank you in advance!
EDIT: I just added an additional if statement if($query3) and it now works.
You are checking for POST variables, but you are getting to edit.php through a GET request. There isn't anything on $_POST. Therefore it drops down to the else of your if block and prints out Testing For Error...
Your script in getting into the else part. That means there nothing it is getting as $_POST['submit']. Make sure that your submit button must have a name attribute as submit.
<input type="submit" name="submit" value="" />
please check what showing in error.log file. You may insert these lines at your edit.php file
error_reporting(E_ALL);
ini_set('display_errors', 1);
to display error.
Replace your else part by this for more detailed mysql errors
else{ echo "Testing For Error...." .mysql_error(); }

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