mbb87 Posted August 7, 2012 Share Posted August 7, 2012 Trying to Update the MySQL Tables but nothing is being Updated, I'm sure I'm just not seeing the tiny issue, would love some help. Thanks So its a trade block for a hockey pool, if you want the player on the trade block then you just check the CHECKBOX in the form and submit and it should change the value in the database to value of "1". FORM: echo "<table border='1'>"; echo "<tr><th>NAME</th> <th>POS</th> <th>BLOCK</th></tr>"; $counter = 1; while($row = mysql_fetch_array( $result )) { echo "<tr><td>"; echo "{$row['f_name']}" . " " . "{$row['l_name']}"; echo "</td><td><input name='pl_id[$counter]' type='hidden' value='{$row['pl_id']}'>"; echo "{$row['pos']}"; echo "</td><td><input name='pos[$counter]' type='hidden' value='{$row['pos']}'>"; echo "<input type='checkbox' name='block[$counter]' size='1' value='1'"; if($row['block'] == '1') { echo "checked='checked'"; } echo "></td></tr>"; $counter++; } echo "</table>"; SUBMIT PHP PAGE: mysql_connect("localhost", "user", "pass") or die(mysql_error()); mysql_select_db("mbbcom1_fantrax") or die(mysql_error()); $i = 1; while ($i < 26) { $block = $_POST['block'][$i]; $pl_id = $_POST['pl_id'][$i]; $query = mysql_query("UPDATE 'players' SET `block` = '$block' WHERE `players`.`pl_id` = '$pl_id'"); mysql_query($query); $i++; } echo mysql_close(); Thank you in advance for any help you can provide me. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/ Share on other sites More sharing options...
Drummin Posted August 7, 2012 Share Posted August 7, 2012 You are executing your query twice. $query = "UPDATE 'players' SET `block` = '$block' WHERE `pl_id` = '$pl_id'"; mysql_query($query) OR DIE .mysql_error(); Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367532 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Nope still doesn't work. No error messages just like before, just doesn't update anything as of now. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367799 Share on other sites More sharing options...
jazzman1 Posted August 8, 2012 Share Posted August 8, 2012 Why do you want to loop the query 26 times Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367801 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Theirs 26 players on each team. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367803 Share on other sites More sharing options...
jazzman1 Posted August 8, 2012 Share Posted August 8, 2012 Okay, but why did you want to update their columns on the same time Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367805 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Here is the page if that help you understand what I want to do. http://www.mbb87.com/fantrax/myblock.php?user=92&inblock=1 Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367808 Share on other sites More sharing options...
Psycho Posted August 8, 2012 Share Posted August 8, 2012 NEVER EVER run queries in loops. They are a huge resource hog and will kill your server. There is ALWAYS a better solution. In this case you need to run only two queries: one to update the record that are unchecked and one to update those that are checked. First off, I would change the form code by removing the $counter and just using the ID of the record as the field index in the input field name. You aren't using the index of the field name for anything anyway. Or you could just remove the named index entirely. echo "<table border='1'>"; echo "<tr><th>NAME</th><th>POS</th><th>BLOCK</th></tr>"; while($row = mysql_fetch_array( $result )) { $checked = ($row['block'] == '1') ? ' checked="checked"' : ''; echo "<tr>\n"; echo " <td>{$row['f_name']} {$row['l_name']}</td>\n"; echo " <td><input name='pl_id[$row['pl_id']]' type='hidden' value='{$row['pl_id']}'>{$row['pos']}</td>\n"; echo " <td>\n"; echo " <input name='pos[$counter]' type='hidden' value='{$row['pos']}'>"; echo " <input type='checkbox' name='block[$counter]' size='1' value='1'{$checked}>\n"; echo " </td>\n"; echo "</tr>"; } echo "</table>"; Then on your processing page you just need to do ONE query like this //Verify the values are valid integers and put in comma separated string $checked_players_list = implode(', ', array_map('int', $_POST['pl_id'])); //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE 'players' SET `block` = `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); EDIT: Corrected one mistake in code above (imploded IDs) and revised to only need ONE query to update all the records! Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367822 Share on other sites More sharing options...
Psycho Posted August 8, 2012 Share Posted August 8, 2012 OK, I just realized that you were using a hidden field for the ID. That's completely unnecessary and would not work with the code I provided. You only need to use the checkbox field in your form. The hidden fields are not needed. Use this for your form along with the processing logic I provided above. echo "<table border='1'>"; echo "<tr><th>NAME</th><th>POS</th><th>BLOCK</th></tr>"; while($row = mysql_fetch_array( $result )) { $checked = ($row['block'] == '1') ? ' checked="checked"' : ''; echo "<tr>\n"; echo " <td>{$row['f_name']} {$row['l_name']}</td>\n"; echo " <td>{$row['pos']}</td>\n"; echo " <td><input type='checkbox' name='pl_id[{$row['pl_id']}]' value='{$row['pl_id']}'{$checked}></td>\n"; echo "</tr>"; } echo "</table>"; Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367825 Share on other sites More sharing options...
Psycho Posted August 8, 2012 Share Posted August 8, 2012 -- DELETED -- Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367826 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Where are you getting "$checked_players_list"? EDIT: Never mind, I see it now but still nothing gets updated. Didnt think it was that confusion, but no one seems to be able to figure it out. I really hope to hearing from you some more. If you have any questions please fire them away. I REALLY WANT TO FIX/GET IT TO WORK THANKS AGAIN FOR YOUR TIME. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367829 Share on other sites More sharing options...
Psycho Posted August 8, 2012 Share Posted August 8, 2012 Where are you getting "$checked_players_list"? EDIT: Never mind, I see it now but still nothing gets updated. Didnt think it was that confusion, but no one seems to be able to figure it out. I really hope to hearing from you some more. If you have any questions please fire them away. I REALLY WANT TO FIX/GET IT TO WORK THANKS AGAIN FOR YOUR TIME. Put some error/debugging handling into your code to see where the problem is. Change the processing code to this: //Verify the values are valid integers and put in comma separated string $checked_players_list = implode(', ', array_map('int', $_POST['pl_id'])); //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE 'players' SET `block` = `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); echo "Debugging info:<br>\n"; echo "POST Data:<pre>" . print_r($_POST, 1) . "</pre>\n"; echo "Checked Players list: {$checked_players_list}<br>\n"; echo "Update Query: {$query}<br>\n"; if($result) { echo "Query executed with " . mysql_affected_rows($result) . " affected rows"; } else { echo "Query Failed. Error " . mysql_error(); } What is output to the page? Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367833 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Debugging info: POST Data: Array ( [pl_id] => Array ( [36] => 36 [31] => 31 [75] => 75 [17] => 17 [187] => 187 [121] => 121 [89] => 89 [42] => 42 [30] => 30 [116] => 116 ) [submit] => submit ) Checked Players list: Update Query: UPDATE 'players' SET `block` = `players`.`pl_id` IN () Query Failed. Error 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 ''players' SET `block` = `players`.`pl_id` IN ()' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367836 Share on other sites More sharing options...
DavidAM Posted August 8, 2012 Share Posted August 8, 2012 Checked Players list: Update Query: UPDATE 'players' SET `block` = `players`.`pl_id` IN () Query Failed. Error 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 ''players' SET `block` = `players`.`pl_id` IN ()' at line 1 Did you read and understand that output? 1) The checked players list is empty. I don't think that "int" is a function. I think that line needs to be $checked_players_list = implode(', ', array_map('intval', $_POST['pl_id'])); 2) Your query is missing a couple of things. It has SET `block` = `players`.`pl_id` IN () but in your earlier posts, it was SET `block` = $block. Then it is missing the "WHERE" -- it should (probably) be $query = "UPDATE 'players' SET `block` = $block WHERE `players`.`pl_id` IN ({$checked_players_list})"; Programming requires some analytical thought. You shouldn't just cut-and-paste suggestions from people without trying to understand what you are pasting. Psycho really gave you the answer, there; it just had a couple of minor issues. If you read his signature you see the disclaimer that any code he provides may not be perfect. The code I provided above may not be perfect. This is not because we are bad programmers. We just don't have the time, and in some cases, the capability, to build a complete environment to fully test your problem. Often we do not have enough information from you to fully build/test the code. We expect you to make an effort to understand and integrate our suggestions. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367842 Share on other sites More sharing options...
mikosiko Posted August 8, 2012 Share Posted August 8, 2012 in addition to all the good pointers already given I'm still seeing the table enclosed in single quotes instead of backtics (or nothing) $query = "UPDATE 'players' SET `block` = $block WHERE `players`.`pl_id` IN ({$checked_players_list})"; Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367844 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 echo "<table border='1'>"; echo "<tr><th>NAME</th><th>POS</th><th>BLOCK</th></tr>"; while($row = mysql_fetch_array( $result )) { $checked = ($row['block'] == '1') ? ' checked="checked"' : ''; echo "<tr>\n"; echo " <td>{$row['f_name']} {$row['l_name']}</td>\n"; echo " <td>{$row['pos']}</td>\n"; echo " <td><input type='checkbox' name='pl_id[{$row['pl_id']}]' value='{$row['pl_id']}'{$checked}></td>\n"; echo "</tr>"; } echo "</table>"; RUNNING IT: mysql_connect("localhost", "username", "password") or die(mysql_error()); mysql_select_db("mbbcom1_fantrax") or die(mysql_error()); //Verify the values are valid integers and put in comma separated string $checked_players_list = implode(', ', array_map('int', $_POST['pl_id'])); //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE 'players' SET `block` = `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); echo "Debugging info:<br>\n"; echo "POST Data:<pre>" . print_r($_POST, 1) . "</pre>\n"; echo "Checked Players list: {$checked_players_list}<br>\n"; echo "Update Query: {$query}<br>\n"; if($result) { echo "Query executed with " . mysql_affected_rows($result) . " affected rows"; } else { echo "Query Failed. Error " . mysql_error(); } where are you guys getting $block ? Im trying to understand what you guys are doing, but not sure how it will update all the players without a loop. you lost me there. (I keep searching online for the solution, not trying to be lazy here and just cut and paste) I REALLY APPRECIATE ALL THE HELP and just trying to understand and get this to work. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367846 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Shouldn't it be something like this? //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE 'players' SET `block` = '1' WHERE `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); Also in the error codes is says '$checked_players_list' is empty no? Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367857 Share on other sites More sharing options...
Drummin Posted August 8, 2012 Share Posted August 8, 2012 You are not reading the replies. 1. Fix the back-ticks issue. $query = "UPDATE `players` SET `block` = `players`.`pl_id` IN ({$checked_players_list})"; 2. Change callback function to interval. $checked_players_list = implode(', ', array_map('intval', $_POST['pl_id'])); Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367858 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 WOW!!! YOU GUYS ARE AMAZING!!! ITs working. I would love to know why though. But i bugged you guys enough, ill try to google it. Thanks again everyone. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367862 Share on other sites More sharing options...
jazzman1 Posted August 8, 2012 Share Posted August 8, 2012 @ mbb87.... CONGRATULATIONS......THOSE GUYS ARE REALLY AMAZING!!! It's a my solution, I found it few months ago and works perfect ( I can not remember the url of the site). You can use CASE operator. // An array containing the category ids as keys and the new positions as values $display_order = array( 1 => 4, 2 => 1, 3 => 2, 4 => 3, 5 => 9, 6 => 5, 7 => 8, 8 => 9 ); $ids = implode(',', array_keys($display_order)); $sql = "UPDATE `tbl_name` SET `id` = CASE id "; foreach ($display_order as $id => $ordinal) { $sql .= sprintf("WHEN %d THEN %d ", $id, $ordinal); } $sql .= "END WHERE `id` IN ($ids)"; $result = mysql_query($sql) or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367866 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Alright so it changes all players that are not in the get query as well, At this point what it does is: Even though the players are not on that users team it changes them back to 0 and removes other users trade block players. Ill try to figure it out. maybe add and other WHERE criteria. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367867 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 So here is the final product for everyone that helped me and if anyone wants a suggestions. Thanks Again!! // Get all the data from the "example" table $result = mysql_query("SELECT * FROM `players` WHERE `own` =$user ORDER BY `players`.`l_name` ASC LIMIT 0 , 30") or die(mysql_error()); ?> <form name="myblock" action="_myblock.php?user=<?php echo $user; ?>" method="POST"> <?php echo "<table border='1'>"; echo "<tr><th>NAME</th><th>POS</th><th>BLOCK</th></tr>"; while($row = mysql_fetch_array( $result )) { $checked = ($row['block'] == '1') ? ' checked="checked"' : ''; echo "<tr>\n"; echo " <td>{$row['f_name']} {$row['l_name']}</td>\n"; echo " <td>{$row['pos']}</td>\n"; echo " <td><input type='checkbox' name='pl_id[{$row['pl_id']}]' value='{$row['pl_id']}'{$checked}></td>\n"; echo "</tr>"; } echo "</table>"; mysql_select_db("mbbcom1_fantrax") or die(mysql_error()); //Verify the values are valid integers and put in comma separated string $checked_players_list = implode(', ', array_map('intval', $_POST['pl_id'])); //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE `players` SET `block` = `players`.`pl_id` IN ({$checked_players_list}) WHERE `own` = $user"; $result = mysql_query($query); Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367869 Share on other sites More sharing options...
Psycho Posted August 8, 2012 Share Posted August 8, 2012 I made some minor errors in the code I provided (I didn't test it), but the logic is sound. Some of the replies since are getting this conversation on the wrong track, so let me elaborate: 1) The checked players list is empty. I don't think that "int" is a function. I think that line needs to be $checked_players_list = implode(', ', array_map('intval', $_POST['pl_id'])); David was 100% correct here - I meant to use 'intval'. Using that will force all the values to be integers to prevent SQL Injection. [ Shouldn't it be something like this? //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE 'players' SET `block` = '1' WHERE `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); As one other contributor noted the table name should be in backticks (or none at all). However, the query is fine. The query I provided doesn't need a WHERE clause and using one would actually make it fail to produce the correct results. I actually tested this logic on a table so I know it works. Let me explain how it works because it is not completely obvious how it works but is actually pretty ingenious in my humble opinion. Here is the query UPDATE `players` SET `block` = `players`.`pl_id` IN (1, 5, 9, 10, 11, 12) The list of IDs will be dynamic based upon those that are checked on the form page. So the query is going to attempt to update all records and set the value of 'block' to `players`.`pl_id` IN (1, 5, 9, 10, 11, 12) What is happening there is that for each record a condition is done to see if that records id is in the list of supplied ids. The result will be true or false. In MySQL a true/false response is a 1 or 0. So, the result is that the block value is set to '1' for all records in the supplied list and '0' for the rest. I originally supplied two queries to do the 0 and 1 separately and then thought of the above approach. After I tested it I modified my post where I provided that code. So, you should only need to make the changes to the array_map() function to use 'intval' and then use backticks around the table name: //Verify the values are valid integers and put in comma separated string $checked_players_list = implode(', ', array_map('intval', $_POST['pl_id'])); //Run update to set blocked status based on whether the player ID was in the submitted list $query = "UPDATE `players` SET `block` = `players`.`pl_id` IN ({$checked_players_list})"; $result = mysql_query($query); echo "Debugging info:<br>\n"; echo "POST Data:<pre>" . print_r($_POST, 1) . "</pre>\n"; echo "Checked Players list: {$checked_players_list}<br>\n"; echo "Update Query: {$query}<br>\n"; if($result) { echo "Query executed with " . mysql_affected_rows($result) . " affected rows"; } else { echo "Query Failed. Error " . mysql_error(); } Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367870 Share on other sites More sharing options...
mbb87 Posted August 8, 2012 Author Share Posted August 8, 2012 Adding the WHERE actually worked fine. Without it, all the rows got updated to 0 if not checked even the ones that i original didn't call (GET) Example: Mikes Team: Karlsson Chara (Checked) Spezza (NOT SHOWING UP ON THE QUERY (GET)) Paul's Team: Ovechkin (CHECKED) Brodeur Well running it without the WHERE would make Chara on the block but would removed all other players that are on other teams from the block by changing the block value to 0 But i thank you so much since without you i would still be scratching my head. Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367872 Share on other sites More sharing options...
Christian F. Posted August 8, 2012 Share Posted August 8, 2012 What is happening there is that for each record a condition is done to see if that records id is in the list of supplied ids. The result will be true or false. In MySQL a true/false response is a 1 or 0. So, the result is that the block value is set to '1' for all records in the supplied list and '0' for the rest. That's quite a neat and interesting trick, didn't realize it would work this way before. Thanks for explaining it! Quote Link to comment https://forums.phpfreaks.com/topic/266773-having-an-issue-with-updating-sql-database-via-php-table/#findComment-1367898 Share on other sites More sharing options...
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