I'm trying to use a variable from a row that is being selected from one table in a SELECT function targetted towards a second table. I tried doing this:
while ($row = mysqli_fetch_assoc($result)) {
$sender = $row['sender_id'];
$sql = "SELECT * FROM users WHERE user_id='$sender'";
$result = mysqli_query($conn, $sql);
while ($row2 = mysqli_fetch_assoc($result)) {
echo $row2['user_uid'];
}
}
I understand that this is not how you set a variable for each row, but I have no clue how it should be done otherwise. At the moment it only displays the username of the sender of the first notification in my inbox database, not the rest.
If anyone knows how to set a variable per row, please let me know. Thank you in advance.
Related
For some reason the first two query's don't show anything. I made it so, that you press a submit button to post your input, and then have it controlled and added to the database. But it doesn't work.
$sql = "SELECT * FROM rubrieklink WHERE rekeningnummer = '".$_POST["rekeningnummer"]."'";
$resultrubrieken = $mysqli->query($sql);
$sql="INSERT INTO betalingen(
bedrijf
,rekeningnummer
,prijs
,rubriek)
VALUES('".$_POST["name"]."'
,'".$_POST["rekeningnummer"]."'
,".$_POST["prijs"]."
,'".$resultrubrieken["rubriek"]."')";
$result = $mysqli->query($sql);
$resultbetalingen = $mysqli->query("SELECT * FROM betalingen");
if ($resultbetalingen->num_rows > 0) {
echo "<center><table><tr><td>bedrijf</td><td>rekeningnummer</td><td>prijs</td><td>rubriek</td></tr>";
while($betalingen = $resultbetalingen->fetch_assoc()){
echo"<tr><td>".$betalingen['bedrijf']."</td><td>".$betalingen['rekeningnummer']."</td><td>".$betalingen['prijs']."</td><td>".$betalingen['rubriek']."</td></tr>";
}
echo"</table>";
}
is rekeningnummer an INTEGER or a string if so you must quote your strings same goes for the insert for example $_POST["name"] is not quoted
PS also why is the INSERT INTO doubled in your query?
I have one database with two tables: "music" and "agenda".
But for some reason once I have queried one table, I cannot perform a similar query on the other table. Or in any case, its variables are empty.
I'd think I could just keep the connection open and perform a second query after the first "while". Like so:
<?php
mysql_connect('localhost', 'root', 'root');
mysql_select_db('erikverwey');
$result = mysql_query("SELECT * FROM agenda ORDER BY date DESC LIMIT 0, 2") or die(mysql_error());
while($row = mysql_fetch_array($result)) {
$count++;
$date[$count] = $row['date'];
$time[$count] = $row['time'];
$place[$count] = $row['place'];
$venue[$count] = $row['venue'];
$who[$count] = $row['who'];
$concert[$count] = $row['concert'];
$urlvenue[$count] = $row['urlvenue'];
}
$result2 = mysql_query("SELECT * FROM music ORDER BY id LIMIT 0, 5") or die(mysql_error());
while($row = mysql_fetch_array($result2)) {
$count++;
$song[$count] = $row['song'];
$artist[$count] = $row['artist'];
$duration[$count] = $row['duration'];
$url[$count] = $row['url'];
}
mysql_close();
?>
But no. In this case, all the variables from the table "music" remain empty.
I've been looking for an answer, but no luck. I'm still new to MySQL, though, so apologies beforehand if this is standard stuff. Thanks!
I found the glitch. Because the counter "$count" was used a second time, it started where it left off and couldn't find any data.
Use a different counter, also in the variables (!), and all is good.
I want use mysql_data_seek with PDO from google search I found that it should looks like this:
$row0 = $result->fetch(PDO::FETCH_ASSOC, PDO::FETCH_ORI_ABS, 0);
however it's not work, what I do wrong?
this is my code:
$query = "SELECT name,age FROM users";
$q = $db->prepare($query);
$q->execute();
$q->setFetchMode(PDO::FETCH_ASSOC);
$arrayData = $q->fetchAll();
foreach ($arrayData as $row){
echo $row['name'] ." ";
echo $row['age'] ."<br>";
}
$result = $q->fetch(PDO::FETCH_OBJ,PDO::FETCH_ORI_ABS,4);
var_dump($result);
I just want get the 5th row in object form from the last run query. I don't want run this query again (as some guys told me) I just want the results from sql buffer.
the var_dump result is: bool(false)
any ideas?
EDIT:
thanks for your answers and sorry but maybe I don't explain myself as well. I like the trick with JSON but the point is that the 5th row is example. I just want use the result of the query from the buffer with PDO exactly as I did it with mysql_data_seek in regular mysql (change the cursor). is it possible? I like all the tricks but that not what I look for.
the PDO 'cursor' default is PDO::CURSOR_FWDONLY that means that cursor can't back to zero like it happens with mysql_data_seek to allow cursor back to zero it necessary define use 'scrollable cursor'
example:
$db->prepare($query, array(PDO::ATTR_CURSOR => PDO::CURSOR_SCROLL));
before use it like this:
$row0 = $result->fetch(PDO::FETCH_ASSOC, PDO::FETCH_ORI_ABS, 0);
$result = $arrayData[4];
is all you need.
If you want the 5th row result you can do like this:
$result = json_decode(json_encode($arrayData[4]), FALSE);
var_dump($result);
or something like this:
$object = new stdClass();
foreach ($array as $key => $value)
{
$object->$key = $value;
}
Just curious! why do you need the object form?
EDIT (this will give the object form of the 5th row):
$index = 0;
$fifthRow = new stdClass();
while($row = $q->fetch())
{
if($index++==4)
$fifthRow = json_decode(json_encode($row), FALSE);
}
You could do it like this:
$c = 1;
$saved=null;
while($row = $q->fetch()){
if($c==4){
$saved = clone $row;
};
$c++;
somethingelse;
}
$saved will then contain the 4th element as an object with almost no extra overhead calculations.
I have this script
$query = "SELECT id,last_name,first_name FROM users WHERE tmima_id='6'";
$result = #mysql_query($query);
while($row = mysql_fetch_array($result))
{
$i = 3;
$emp_id = $row['id'];
$cell = 'A'.$i;
$objPHPExcel->setActiveSheetIndex(0)
->setCellValue($cell, $row['last_name']. $row['first_name']);
$i++;
}
But in the .xls file it prints only one user. Why id doesnt print all of the users ? W
Thanks in advance.
I make the change you said with $sheet
$query = "SELECT id,last_name,first_name FROM users WHERE tmima_id='6'";
$result = #mysql_query($query);
while($row = mysql_fetch_array($result))
{
$i = 3;
$emp_id = $row['id'];
$cell = 'A'.$i;
$sheet->setCellValue($cell, $row['last_name']. $row['first_name']);
$i++;
}
But it still prints out only one record. And yes when i run the query in phpmyadmin it returns more than one record.
How can i print out data from mySql table.. What is going wrong ?
I am pretty sure it is because you are using a unique identifier (WHERE tmima_id='6'). It is only finding the results for that one unique identifier and displaying that. Hope this helps.
$i is being reset to row 3 every loop. Set $i=3; before the while loop, not inside it.
Been staring at this all day and can't seem to figure out why my update statement fails to update the field 'image_filename':
$fileName = $_FILES['image_filename'];
if($fileName["name"] <> ""){
$imageFile = $fileName['name'];
$destination = "../../../../assets/resources/images/".$fileName['name'];
move_uploaded_file($fileName['name'], $destination);
}
$updateSQL = sprintf("UPDATE content SET image_filename='$imageFile' WHERE id=%s",
GetSQLValueString($_POST['resource_id'], "int"));
mysql_select_db($database_conn_talent, $conn_talent);
$Result1 = mysql_query($updateSQL, $conn_talent) or die(mysql_error());
Can a SQL pro tell me what I"m missing? Much thanks in advance for your feedback!
You appear to be building a query, but never executing it. Also, Drupal'll handle all the sprintfing for you, if you let it.
$query = "UPDATE content SET image_filename='$imageFile' WHERE id=%i";
db_query($query, $_POST['resource_id']);