How to loop php using ref data from mysql? - php

How to loop php using ref data from mysql ?
work flow :
Loop If column 'xxx' == ''
{
$sql = "SELECT * FROM table WHERE xxx == '' order by id desc";
$result = mysql_query($sql);
$datas=mysql_fetch_array($result);{
$id = stripslashes(str_replace('\r\n', '<br>',($datas['id'])));
}
$sql = "UPDATE table SET xxx = 'test' WHERE id = '$id' ";
$dbQuery = mysql_query($sql);'
}

ypur query is confusing. why would you store ids with html in them? regardless, this should work
$sql = "SELECT * FROM table WHERE xxx != '' "; //ORDER BY will not work since ids are not int
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result){
$id = stripslashes(str_replace('\r\n', '<br>',($row['id'])));
$sql = "UPDATE table SET xxx = 'test' WHERE id = '$id' ";
$dbQuery = mysql_query($sql);'
}

Related

Array to string conversion in mysql while loop

Following is my code,
$result1 = "SELECT emp_id FROM employee where manager_id=".$userID;
$array = mysql_query($result1);
$cnt = 0;
while ($row = mysql_fetch_array($array)) {
"emp_id: " . $row[0];
$myArrayOfemp_id[$cnt] = $row[0];
$cnt++;
}
var_dump($myArrayOfemp_id);
$sql = "SELECT emp_id FROM emp_leaves WHERE emp_id='$myArrayOfemp_id' ORDER BY apply_date DESC";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
When I'am trying to use $myArrayOfemp_id variable in $sql query, It shows that error:
Array to string conversion in..
How can I fix it?
You are trying to convert an array into a string in the following line:
$sql = "SELECT emp_id FROM emp_leaves
WHERE emp_id='$myArrayOfemp_id' ORDER BY apply_date DESC";
$myArrayOfemp_id is an array. That previous line of code should be changed to:
$sql = "SELECT emp_id FROM emp_leaves
WHERE emp_id={$myArrayOfemp_id[0]} ORDER BY apply_date DESC";
I placed 0 inside {$myArrayOfemp_id[0]} because I'm not sure what value want to use that is inside the array.
Edited:
After discussing what the user wanted in the question, it seems the user wanted to use all the values inside the array in the sql statement, so here is a solution for that specific case:
$sql = "SELECT emp_id FROM emp_leaves
WHERE ";
foreach ($myArrayOfemp_id as $value)
{
$sql .= " emp_id={$value) || ";
}
$sql .= "1=2";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
$sql = "SELECT emp_id FROM emp_leaves WHERE emp_id in
(SELECT GROUP_CONCAT(emp_id) FROM employee where manager_id=".$userID.")
ORDER BY apply_date DESC";
$result = mysql_query($sql);
$total_results = mysql_num_rows($result);
just change your query like above might solve your problem.
you can remove following code now. :)
$result1 = "SELECT emp_id FROM employee where manager_id=".$userID;
$array = mysql_query($result1);
$cnt = 0;
while ($row = mysql_fetch_array($array)) {
"emp_id: " . $row[0];
$myArrayOfemp_id[$cnt] = $row[0];
$cnt++;
}
var_dump($myArrayOfemp_id);

PHP not displaying result from MYSQL query

I have the following code
$sql = "SET #uid := (SELECT ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);";
$sql = "UPDATE channels SET Used = 1 WHERE ID = #uid;";
$sql = "SELECT * FROM channels WHERE ID IN = #uid;";
$result = mysqli_multi_query($conn, $sql)
or die( mysqli_error($sql) );
if (mysqli_num_rows($result) > 0) {
$text = '';
while($row = mysqli_fetch_assoc($result)) {
$Channel_Location = $row['Channel_Location'];
$text = $text . $Channel_Location;
}
}
Now the issue i'm having is the php isnt displaying the result returned by the MYSQL query which is stored in a session later on in the code to be displayed on a dummy page it comes up with the following error
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result
The my SQL query does exactly what I need it to I just need to, so I don't really want to change it. I just need some advice on how i'd get the PHP to echo the #uid
is there anyone willing to help me solve the issue? if so thankyou.
You have 3 queries in your $sql so you should use multi_query function
http://php.net/manual/en/mysqli.multi-query.php
And you can change your first query to:
SET #uid = 0;
SELECT #uid := ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);
Update You can try this fragment of your code modified with all commented improvements.
$sql = 'SET #uid = 0;';
$sql .= 'SELECT #uid:= ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);';
$sql .= 'UPDATE channels SET Used = 1 WHERE ID = #uid;';
$sql .= 'SELECT * FROM channels WHERE ID IN = #uid;';
if (mysqli_multi_query($conn, $sql)) {
do {
$result = mysqli_store_result($conn);
} while(mysqli_next_result($conn));
if (mysqli_num_rows($result) > 0) {
$text = '';
while($row = mysqli_fetch_assoc($result)) {
$Channel_Location = $row['Channel_Location'];
$text = $text . $Channel_Location;
}
}
} else {
die( mysqli_error($conn) );
}

Update rows of a SQL table

I have a table with a column called 'status'. The defaut value of 'status' is 0. I want to update the value to '1' after using it.
I basically want to check if the status is 0, if it is, do an operation and then change the value to 1.
Here is the code. All works perfectly except that the value of 0 is not changed to 1.
I am a novice so maybe is a very basic mistake :(
<?php
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$sql = "UPDATE summonerid SET status='1' WHERE status='0';"; // THIS IS THE PART THAT DOES NOT WORK WELL
}
}
}
?>
Any help would be highly appreciated
Try this.. you are not executing the sql statements
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$result1 = $conn->query($sql);
$sql = "UPDATE summonerid SET status='1' WHERE status='0';";
$result1 = $conn->query($sql);
}
}
}
?>

Random row selection using MySQL returns NULL

I am trying to get a random row from MySQL table but all three attemps:
$query = "SELECT cid FROM table LIMIT 1 OFFSET ".rand(1,$num_rows);
$query = "SELECT cid FROM table OFFSET RANDOM() * (SELECT COUNT(*) FROM table) LIMIT 1";
$query = "SELECT * FROM table ORDER BY RAND() LIMIT 1";
give a NULL result in mysql_query($query).
Higher up my PHP code I obtain a row from the same table OK by specifying WHERE, so I don't understand why I can't retrieve a random one.
Here is the code snippet:
$query = "SELECT uid,clu FROM uable WHERE un = '$un'";
$result = mysql_query($query) or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
$resultid = mysql_fetch_assoc($result);
$uid = $resultid['uid'];
file_put_contents('debugging.txt',__LINE__.' - $uid = '.var_export($uid,true).PHP_EOL,FILE_APPEND);
$query = "SELECT * FROM table WHERE uid = $uid AND cn = '$cn'";
$result = mysql_query($query) or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
$cr = mysql_fetch_assoc($result);
$cid= $cr['cid'];
file_put_contents('debugging.txt',__LINE__.' - $cid= '.var_export($cid,true).PHP_EOL,FILE_APPEND);
$query = "SELECT * FROM fable WHERE cid= '$cid'";
$result = mysql_query($query) or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
file_put_contents('debugging.txt',__LINE__.' - $result = '.var_export($result,true).PHP_EOL,FILE_APPEND);
$fr = mysql_fetch_assoc($result);
file_put_contents('debugging.txt',__LINE__.' - $fr = '.var_export($fr,true).PHP_EOL,FILE_APPEND);
echo '<form action="'.$_SERVER['PHP_SELF'].’" method="post">';
if (!$fr) {
$o= $cn;
while ($o= $cn) {
// $ac = mysql_query("SELECT * FROM table") or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
// $num_rows = mysql_num_rows($ac);
//file_put_contents('debugging.txt',__LINE__.' - $num_rows = '.$num_rows.PHP_EOL,FILE_APPEND);
// --$num_rows;
// $query = "SELECT cid FROM table LIMIT 1 OFFSET ".rand(1,$num_rows);
$query = "SELECT cid FROM table OFFSET RANDOM() * (SELECT COUNT(*) FROM table) LIMIT 1";
// $query = "SELECT * FROM table ORDER BY RAND() LIMIT 1";
$resultid = mysql_query($query) or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
$opr = mysql_fetch_assoc($resultid);
$o= $opr['cn'];
}
file_put_contents('debugging.txt',__LINE__.' - $query = '.$query.PHP_EOL,FILE_APPEND);
file_put_contents('debugging.txt',__LINE__.' - $resultid = '.var_export($resultid,true).PHP_EOL,FILE_APPEND);
file_put_contents('debugging.txt',__LINE__.' - $op[\'cid\'] = '.$op['cid'].PHP_EOL,FILE_APPEND);
$query = "SELECT * FROM table WHERE cid= ".$op;
$result = mysql_query($query) or die(sqlerror(__LINE__,mysql_errno(),mysql_error()));
$opr = mysql_fetch_assoc($opr);
$o= $opr['cn'];
$od= $opr['description'];
echo '<p>'.$op;
if ($od<> '') {
echo ','.$odesc;
}
echo '</p>';
echo '<input type="submit" name="continue" id="continue" value="Continue">';
} else {
echo '<p>'.$fr['p'].'</p>';
echo '<input type="submit" name="continue" id="continue" value="Continue">';
}
echo '</form>';
The resulting debugging.txt:
24 - $uid = '4'
29 - $cid = '21'
32 - $result = NULL
34 - $fr = false
These queries look OK, but I think you're starting at the wrong place. When you're uncertain how to frame something in SQL, open up a SQL client like SequelPro or Navicat and try writing a few queries by hand until you get the result you want. (Also this gives you a chance to double-check the contents of relevant tables and ensure the expected data are there.) Then you can go back into the PHP with full confidence that the SQL code is correct, so if there's a problem it must be with the PHP (either the variables you inject into a Mysql statement, or the way you call that statement).

MYSQL UPDATE in PHP using AES_ENCRYPT

I am trying to do a basic mysql update using AES_ENCRYPT in PHP but I cant get it work.
Here is my full code:
$key = "bac09c63f34c9845c707228b20cac5e0";
$query = " SELECT id, aes_decrypt(Column1, '$key') AS Column1, aes_decrypt(Column2, '$key') AS Column2 FROM parent WHERE Request = '{$Request}' ORDER BY ID ASC;";
$resultSet = mysql_query($query, $DB);
while ($row = mysql_fetch_array($resultSet)) {
$id = $row['ID'];
$rows[] = $row["Request"];
$Column1 = $row["Column1"];
$Column2 = $row["Column2"];
$SQL = "UPDATE parent SET Column1 = AES_ENCRYPT('$Column1','$key'), Column2 = AES_ENCRYPT('$Column2','$key') WHERE Parent_ID = '$id';";
if (!mysql_query($SQL, $DB))
die("Query Failed $SQL");
}
PHP error log is fine, there is no error. Only this:
Query Failed
UPDATE parent SET Column1 = AES_ENCRYPT('722225374673255299521908919676768...etc','bac09c63f34c9845c707228b20cac5e0')
Use this query -
UPDATE parent SET Column1 = AES_ENCRYPT('$Column1','$key'), Column2 = AES_ENCRYPT('$Column2','$key') WHERE Parent_ID = $id;

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