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PHP - Help With Mysql Update Query...cant See What The Heck Is Going Wrong.
Ok, I'm going start off simple. If I have to provide more code I will.
I am doing an update on a table called countries. Yet my query just will not update the db. Is there anything wrong with this query? mysql_query("UPDATE countries SET country_id = '{$_POST['update_value']}' WHERE country_id = '{$_POST['original_html']}'") or die(mysql_error()); Similar TutorialsI have a value that in parts of my code i need to update. Its a number value, stored as TEXT. which i need to add numbers to it, at the moment I am using this to get the current value: Code: [Select] $sql2="SELECT * FROM weekly LIMIT 1"; $result2=mysql_query($sql2); $row2 = mysql_fetch_array( $result2 ); $rollingjackpot = $row2['rollingjackpot']; then i am adding the value to the varible, then using UPDATE to update it again, cant this be done in a simple query? something like: Code: [Select] $query = mysql_query("UPDATE weekly SET `rollingjackpot` + $amount WHERE `id` = '1'") or die(mysql_error()); This isn't the entire code just enough to see what I'm trying to do. Everything was working until I added the mysql update query in the if statement. Is this possible or am I doing something wrong? When I run the script it just echos "No results found" twice as $num_results = 2. Code: [Select] <?php include("../includes/connect.php"); $query = "SELECT ........ "; $result = $db->query($query); $num_results = $result->num_rows; if ($num_results == 0) { exit; } else { $i=0; while ($i < $num_results) { $row = $result->fetch_assoc(); $id = $row['id']; if ($expiration_date > $today) { ### EMAIL CODE HERE ### $update = "UPDATE model SET reminder_sent = '1' WHERE id = '$id' "; $result_2 = $db->query($update); $i++; } else { echo "No results found."; $i++; } } } ?> Hi guys need some help.I created a simple Update Customer Details page Where you Enter the Customer ID ,The Customer Details Get Displayed inside a Form and you make changes within the Form.But the update query I wrote is not working as it should be.It Executes the Query ,when I hardcode the Customer ID inside the Update query but fails when I need to Enter the Customer ID directly from the form using post.In This Code I tried to Update only the Customer's Firstname.Thanks Here Is The Code Code: [Select] <?php $db=mysql_connect("localhost","root","") or die('Unable to Connect To Database.Please check the Database Parameters'); mysql_select_db('ecommerce') or die(mysql_error()); if(isset($_POST['enterid'])) { $_POST['inputid']; } $query="SELECT * FROM customer WHERE Cid='$_POST[inputid]'"; $result=mysql_query($query) or die(mysql_error()); $row=mysql_fetch_array($result); $new_cid=$row['Cid']; $new_firstname=$row['Cfname']; $new_lastname=$row['Clname']; $new_email=$row['Email_id']; $new_address=$row['Address']; $new_pincode=$row['Pincode']; $new_payment=$row['Mode_of_payment']; $new_city=$row['City']; $new_state=$row['State']; $new_phone=$row['Phone']; $html1=<<<HTML1 <form action="updatecustomer.php" method="post"> <p class="inputidentifier">Please Enter The Customer's ID</p><input type="text" style="width:375px;height:40px;font-size:30px;" name="inputid" size="20"><br><br> <input type="submit" name="enterid" style="height:40px"> HTML1; print($new_cid); if(isset($_POST['update'])) { $query1="UPDATE customer SET Cfname='$_POST[firstname]' WHERE Cid='$_POST[inputid]'"; mysql_query($query1) or die(mysql_error()); if(mysql_affected_rows()==1) print("Query sucessful"); else print("Something went wrong"); } ?> <html> <head> <style type="text/css"> .inputtext {width:300px; height:40px;font-size:30px;} .inputidentifier{font-size:25px;font-family:"Arial"} .h1type{font-family:"Arial"} </style> <title>Test</title> </head> <body> <?php print($html1); ?> <h1 align="center" class="h1type">Update Customer Details</h1> <form action="updatecustomer.php" method="POST"> <table align="center" cellspacing="10" cellpadding="10" border="0" width="60%"> <tr> <td align="right" class="inputidentifier">First Name</td> <td align="left"><input type="text" class="inputtext" name="firstname" placeholder="eg:Kevin" value="<?php print($new_firstname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Last Name</td> <td><input type="text" class="inputtext" name="lastname" placeholder="eg:Aloysius" value="<?php print($new_lastname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">E-mail</td> <td align="left"><input type="email" style="width:500" class="inputtext" name="email" placeholder="yourname@email.com" value="<?php print($new_email) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Phone Number</td> <td align="left"><input type="text" class="inputtext" name="phone" placeholder="How Do We Call You?" value="<?php print($new_phone) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Address</td> <td><textarea style="width:500;height:150" wrap="virtual" class="inputtext" name="address" placeholder="Where is your Crib?"><?php print($new_address) ?></textarea></td> </tr> <tr> <td align="right" class="inputidentifier">State</td> <td align="left"><input type="text" style="width:500" class="inputtext" name="state" placeholder="State" value="<?php print($new_state) ?>"> </tr> <tr> <td align="right" class="inputidentifier">City</td> <td align="left"><input type="text" class="inputtext" name="city" placeholder="City" value="<?php print($new_city) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Pin Code</td> <td><input type="text" class="inputtext" name="pincode" placeholder="Mulund 400080" maxlength="6" value="<?php print($new_pincode) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">How do Pay for your Bling?</td> <td align="left"> <input type="text" class="inputtext" name="payment" value="<?php print($new_payment)?>"> </tr> <tr> <td></td> <td><input type="submit" name="update" value="Update!" style="width:100px;height:60px;"></td> </tr> </table> </form> </body> </html> I made a small editing system for my news page, and I need to update three columns within my table "announcements" in the database. I tried a method of updating all of them with one MySQL query instead of using three as it just isn't neat. I've searched several methods via google and I've tried all of them, but just can't seem to get it to work. Is this MySQL query correct? mysql_query("UPDATE announcements SET title = {$title} WHERE id = '$id', content = {$content} WHERE id = '$id', lastmodified = ". date('M-d-Y') ." WHERE id = '$id'"); I have the following PHP script to update two time/date fields in the database. When i run this the fields are not updated. Can anyone see where i m going wrong. <?php $con = mysql_connect("localhost","dbname","dbpassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_select_db("my_db", $con); mysql_query("UPDATE msm_content SET created = '2011-01-02 00:00:00', modified = '2011-01-01 00:00:00'"); echo 'Query Updated successfully'; mysql_close($con); ?> Your guidance is much appreciated. Hi there, I'm having a problem with updating a record with an UPDATE mysql query and then following that query with a SELECT query to get those values just updated. This is what I'm trying to do...I'd like a member to be able to complete a recommended task and upon doing so, go to a page in their back office where they can check off that task as "Completed". This completed task would be recorded in their member record in our database so that when they return to this list, it will remain as "Completed". I'm providing the member with a submit button that will call the same page and then update depending on which task is clicked as complete. Here is my code: Code: [Select] $memberid = $_SESSION['member']; // Check if form has been submitted if(isset($_POST['task_done']) && $_POST['task_submit'] == 'submitted') { $taskvalue = $_POST['task_value']; $query = "UPDATE membertable SET $taskvalue = 'done' WHERE id = $memberid"; $result = mysqli_query($dbc, $query); } $query ="SELECT task1, task2, task3 FROM membertable WHERE id = $memberid"; $result = mysqli_query($dbc, $query); $row = mysqli_fetch_array($result, MYSQLI_ASSOC); $_SESSION['task1'] = $row['task1']; $_SESSION['task2'] = $row['task2']; $_SESSION['task3'] = $row['task3']; ?> <h4>Task List</h4> <table> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task1" /> <input type="hidden" name="task_submit" value="submitted" /> </form> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task2" /> <input type="hidden" name="task_submit" value="submitted" /> </form> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task3" /> <input type="hidden" name="task_submit" value="submitted" /> </form> </table> The problem that I am having is that the database is not updated with the value "done" but after submission, the screen displays "Completed" instead of "Mark As Completed". So the value is being picked up as "done", but that is why I have the SELECT after the UPDATES, so that there is always a current value for whether a task is done or not. Then I refresh and the screen returns the button to Mark As Complete. Also, when I try marking all three tasks as, sometimes all three are updated, sometimes only one or two and again, I leave the page or refresh and the "Marked As Completed" buttons come back. Bizarre. If anyone can tell me where my logic is going wrong, I would appreciate it. Hello, I am trying to pick up php again and just exercising my skills. So I have it so that it fills my form with the values of what I want to edit, and when I click the edit button, it doesn't edit any of the information. When I echo out $result, I get a MYSQL query string that has the same values as the table, so its not getting the new values that are edited. <?php @mysql_connect('localhost', 'root', '') or die("Could not connect to Mysql Server. " . mysql_error()); @mysql_select_db('tutorials') or die("Could not connect to Database. " . mysql_error()); if(isset($_GET['edit'])) { $id = $_GET['edit']; $query = "SELECT `username`, `password` FROM `users` WHERE `id` = '$id'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $name = $row['username']; $password = $row['password']; } if(isset($_POST['edit'])) { $id = $_GET['edit']; $query = "UPDATE `users` SET `username` = '$name', `password` = '$password' WHERE `id` = '$id'"; $result = mysql_query($query); echo $query; if(!$result) { echo mysql_error(); }else{ echo 'updated post'; } } ?> <form method="POST" action="" > <input type="text" name="name" value="<?php echo $name; ?>" /> First name <br /> <input type="text" name="password" value="<?php echo $password; ?>" /> Last name <br /> <input type="submit" name="edit" value="edit" /> </form> I believe it has something to do with the values of $name and $password in the form conflicting with the first if isset and the second if isset. Thanks for any help possible I've messed around with this for three hours now, and it's driving me batty. I'm running the following Query: mysql_select_db("myDB", $con); $query = "UPDATE mytable SET Name = '$Name', Address = '$Address', City = '$City', State = '$State', Zip = '$Zip', Phone = '$Phone', Website = '$Website', Type = '$Type' WHERE id = '$ID'"; The problem is that it's not updating my database. I've tried everything. I've forced the page to Echo out the variables so that I know they're right. I've run (successful) Delete queries to make sure I'm connecting to the right table. I've forced it to spit out the variables at the end of the Update. Everything I've tried works beautifully... except the Update itself. It doesn't error out or anything- it acts like it's Updating, but the data never changes. I can add to the database and delete from it, but it refuses to let me Update anything. I've looked all over the net, and tried every variation I can find. Nothing works. I also tried this: $query = "UPDATE `testbed` SET `Name`=[$Name],`Address`=[$Address],`City`=[$City],`State`=[$State],`Zip`=[$Zip],`Type`=[$Type],`Phone`=[&Phone],`website`=[$Website] WHERE `ID`=[$ID]"; And $result = mysql_query("UPDATE mytable SET Name = '$Name', Address = '$Address', City = '$City', State = '$State', Zip = '$Zip', Phone = '$Phone', Website = '$Website', Type = '$Type' WHERE id = '$ID'"); None of them do anything at all. Hi all, I have the following MySQL insert query: Code: [Select] $insert= mysql_query ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')"); where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values. Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues. So something like this: Code: [Select] $update = mysql_query ("UPDATE tablename SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."') WHERE .... "); Is this possible? If so, what is the proper syntax? Thanks! Hello all,
Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).
The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).
Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:
In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).
I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.
Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.
So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:
1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red.
2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red.
3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.
Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.
Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.
Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.
Cheers guys!
case "update-delete":
if(isset($_POST['highlight-purchased']))
{
// ****** Setup customized query to obtain only items that are checked ******
$sql = "SELECT * FROM shoplist WHERE";
for($i=0; $i < count($_POST['checkboxes']); $i++)
{
$sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or";
}
$sql= rtrim($sql, "or");
$statement = $conn->prepare($sql);
$statement->execute();
// *** fetch results for all checked items (1st query) *** //
$result = $statement->fetchAll();
$statement->closeCursor();
// Setup query that will change the purchased flag to "N", if it's currently set to "Y"
$sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'";
// Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL
$sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL";
$statementSetToN = $conn->prepare($sqlSetToN);
$statementSetToY = $conn->prepare($sqlSetToY);
/*** loop through the results/collection of checked items ***/
foreach($result as $row)
{
if ($row["purchased"] != "Y")
{
// *** fetch one row at a time pertaining to the 2nd query *** //
$resultSetToY = $statementSetToY->fetch();
foreach($resultSetToY as $row)
{
$statementSetToY->execute();
}
}
else
{
// *** fetch one row at a time pertaining to the 2nd query *** //
$resultSetToN = $statementSetToN->fetch();
foreach($resultSetToN as $row)
{
$statementSetToN->execute();
}
}
}
break;
}
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I am querying... $sql = "SELECT `messages_inbox`.`message_id`, `users`.`firstname`, `users`.`lastname`, `users`.`username` AS `from`, '${user_info['username']}' AS `to`, `subject`, LENGTH(`files`) AS `len`, 'inbox' AS `box`, DATE_FORMAT(`messages_inbox`.`time` ,'%T %D-%M-%Y') AS `time` "; $sql .= "FROM `messages_inbox` INNER JOIN `users` ON `messages_inbox`.`from_id` = `users`.`id` WHERE `to_id` = ${user_info['uid']} AND `messages_inbox`.`deleted` = 0 ORDER BY `messages_inbox`.`message_id` DESC"; and I am trying to output $displayName = ucwords("${message['firstname']} ${message['lastname']}"); by using $messages = pm_fetch_all($_GET['box']); I know my fetch works but for some reason firstname and lastname are only returning the logged in users first name and last name, not the person who sent the message. I created this code to upload a member's main picture on his member page on website. I'll only include the query part of the code since that's what is relevant to my problem. The idea is basically to upload a new picture onto the database if no picture already exists for that member and display the picture on the page. If a picture already exists, then the script replaces the old picture with the new one upon upload. But for whatever reason I don't understand, when I try to replace the old pic, it gets inserted in a new row on the database instead of replacing the old row, and the new pic gets displayed on the web page alongside the old. Code: [Select] $query = "SELECT username FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main'"; $result = @mysql_query($query); $num = @mysql_num_rows($result); if ($num> 0) { //Update the image $update = mysql_query("UPDATE images SET image = '" . $image['name'] . "' WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main'"); $_SESSION['error'] = "File updated successfully."; //really should be session success message. header("Location: member.php"); exit; } else { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); $_SESSION['error'] = "File uploaded succussfully."; //really should be session success message. header("Location: member.php"); } So can anyone tell me what the problem is? Could the fact that my insert script actually uploads the image onto a folder on my server and only stores the path name in the database have anything to contribute to the mixup? Appreciate your responses in advance. Can anyone post a generic update function to update mysql table. The manual approach: update $tablename set $column1='a', $column2='b' where $id=$value; Hi guys, I have an update page where users can update their information, i have a different update page which works fine with almost the same code, but this one wont update anything. it does show the data but it wont update it, can u help please? <?php session_start(); include ("../../global.php"); //welcome messaage $username=$_SESSION['username']; echo "$username"; $query=mysql_query("SELECT id FROM users WHERE username='$username'"); while($row = mysql_fetch_assoc($query)) { $user_id = $row['id']; } $ref=$_GET['reference']; if (isset($_POST['register']) && $_POST['register']){ $title = addslashes(strip_tags($_POST['title'])); $update=mysql_query("UPDATE msg SET title ='$title' WHERE reference='$ref'"); } ?> <html> <head> </head> <body> <form action='edit-msgs.php' method='POST' enctype='multipart/form-data'> Title:<br /> <input type='text' name='title' id='title' value='<?php $getdata = "SELECT title FROM msg WHERE reference='$ref' AND referal_id='$user_id'"; $result = mysql_query($getdata); $row = mysql_fetch_assoc($result); echo $row['title'];?>'><p /> <input type='submit' name='register' value='Update'> </form> </body> </html> As you can see, this array is setting each colorvalue to each idvalue. Code: [Select] <?php include("config.php"); include("db.php"); $arrid=$_POST['id']; $arrcolor=$_POST['color']; foreach ($arrid as $idvalue) { foreach ($arrcolor as $colorvalue) { $sql = "UPDATE colors SET color='$colorvalue' WHERE id = '$idvalue'"; mysql_query($sql) or die("Error: ".mysql_error()); echo $sql; } } header("Location: " . $config_basedir . "adminhome.php"); ?> Hi
I just finished this tutorial('http://www.startutor...torial-part-3/)
and everything was working fine until I decided to add last names to the application.
I got everything working on all the other pages except the Update page.
This is my php code.
$q = "SELECT `users.id`, `users.username`, `users.firstname`, `users.lastname`, `users.accounttype`, `companies.companyid`, `companies.companyname`, `companies.companyoccupation`, `companies.country`, `companies.state`, `companies.city`, `companies.industry` FROM `users`, `companies` WHERE `users.id` = `companies.companyid`"; Error: Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in /home/www-data/mysite.com/fresh.php on line 25 $sql_form_union = mysql_query("SELECT id, base FROM Match_1v1 WHERE show_id='$show_id' ORDER BY ordered UNION SELECT id, base FROM Segment_1 WHERE show_id='$show_id' ORDER BY ordered"); while ($form_union = mysql_fetch_array($sql_form_union)){ } Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 |
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