I'm having problems with my php script.
Im settings these variables:
$v1 = mysql_real_escape_string($_POST["v1"]);
$v2 = mysql_real_escape_string($_POST["v2"]);
$v3 = mysql_real_escape_string($_POST["v3"]);
$v4 = mysql_real_escape_string($_POST["v4"]);
I want these values to be updated to the row of my db where id = 1 every time (the row already exists and just need to be updated).
should I then insert or update the row?
I've tried this without success:
$sql = "INSERT INTO table1 (v1, v2, v3, v4) WHERE id = 1";
$sql .= "VALUES ('$v1', '$v2', '$v3', '$v4')";
Use an UPDATE rather than an INSERT.
Try this:
UPDATE table1 set v1 = '$v1', v2 = '$v2', v3 = '$v3', v4 = '$v4' WHERE id = 1
Related
So this is the code I have so far, I'm leaving out the code I used to connect to the database since that's not important and isn't the problem.
$id = strip_tags(mysql_real_escape_string($_GET['id']));
$sql_value = "SELECT value FROM table";
$sql = mysql_query("UPDATE table SET value='[idk what to do here]' WHERE id='$id'");
so the $id selects the id from the url or API and the $sql_value selects the values from the table.
I want the value in the same row as the id specified to increment by 1
E.G.
id = 0, value = 0;
id = 1, value = 0;
id = 2, value = 0;
id = 3, value - 0;
If in the API I type: "id=2"
I want the PHP script to increment the corresponding "value" by 1
E.G.
id = 0, value = 0;
id = 1, value = 0;
id = 2, value = 1;
id = 3, value - 0;
Assuming that your value field is an INT. You can do that by:
"UPDATE table SET value=value+1 WHERE id='$id'"
But i strongly recommend you to take a look at mysqli or PDO and start to make prepared statements to handle the data. See more here:
http://php.net/manual/en/mysqli.quickstart.prepared-statements.php
http://php.net/manual/en/pdo.prepare.php
Use:
$sql = mysql_query("UPDATE table SET value=value+1 WHERE id='$id'");
As long as the value is an int this will work:
UPDATE table SET value=value + 1 WHERE id='$id'
UPDATE tablename t SET value = ( t.value+1 ) WHERE t.id = '$id'
Do you want?
"UPDATE table SET value=value+1 WHERE id='$id'"???
I am trying to issue a mysqli_multi_query, in which my querys are named $query & $query2. Query 1 is a seperate table from query 2. This is a sample of how the code syntax looks like:
$query1 = "INSERT INTO invoices (`id`,`c`) VALUES (NULL, '$client_id')";
$query2 = "UPDATE `customers` SET `a` = `$a`,`b` = `$b` WHERE `customers.id` = $client_id";
the invoices.client_id is the same as the customers.id, and I only want to update customers.id that matches the invoice client_id.
For some odd reason, everything is updated fine into my invoices, but not into my customers. Is my syntax correct?
after our discussion in the chat we figured out the following misstakes:
Your code:
$query2 = "UPDATE customers SET alarmcode = $alarmcode, garagecode = $garagecode, gatecode = $gatecode, liason = $liason, lphone = $lphone WHERE customers.id = '$client_id'";
table-def:
So the problem was not correct encapsulating of strings in the sql statement.
corrected statement was:
$query2 = "UPDATE customers SET alarmcode = '$alarmcode', garagecode = '$garagecode', gatecode = '$gatecode', liason = '$liason', lphone = '$lphone' WHERE id = $client_id";
I am trying to update a patient's record its saying its complete but its not updating phpmyadmin. When I press save and update button it shows Save was successful. Any Ideas ?
<?php
include 'connect.php';
$id1 = $_POST['PatientID']; //Text box the user searches in
mysql_query("UPDATE PatientRecords SET
PatientID = '".$_POST['PatientID']."',
FirstName = '".$_POST['FirstName']."',
LastName = '".$_POST['LastName']."',
DOB = '".$_POST['DOB']."',
IDNumber1 = '".$_POST['IDNumber1']."',
Medication1 = '".$_POST['Medication1']."',
Medication1Dosage = '".$_POST['Medication1Dosage']."',
IDNumber2 = '".$_POST['IDNumber2']."',
Medication2 = '".$_POST['Medication2']."',
Medication2Dosage = '".$_POST['Medication2Dosage']."',
IDNumber3 = '".$_POST['IDNumber3']."',
Medication3 = '".$_POST['Medication3']."',
Medication3Dosage = '".$_POST['Medication3Dosage']."',
MedicalNotes = '".$_POST['MedicalNotes']."'
WHERE PatientID = '$id1');
echo"Patient Information has been updated successfully";
mysql_close($con);
?>
Your query is wrong to begin with. If you're UPDATING an existing row, leave the VALUES (... out and end the query after WHERE PatientID = '$id1'. If you're INSERTING a new row, use INSERT table (column, column, ...) VALUES (value, value, ...) and don't use WHERE.
I'm trying to get php to update a MySQL table using UPDATE statement but it simply won't work. Here's the code I wrote:
$add = "1";
$counter=mysql_query("SELECT * FROM frases WHERE id = '".$id."'");
while ($ntcounter=mysql_fetch_array($counter)) {
mysql_query("UPDATE frases SET count = '".$ntcounter[count]+$add."' WHERE id = '".$id);
}
As you can see, I am basically trying to update the SQL record to keep track of how many times a specific content ID was visited.
Thanks!
Use an alias in your SQL query (It is not mandatory, but it makes the query much more readable.)
SELECT * as count FROM frases WHERE id = '".$id."'"
And you can now access to your variable
$ntcounter['count']
So the result :
$add = "1";
$id = (int)$id
$counter = mysql_query("SELECT * as count FROM frases WHERE id = '".$id."'");
while ($ntcounter = mysql_fetch_assoc($counter)) {
mysql_query("UPDATE frases SET count = '".($ntcounter['count']+$add)."' WHERE id = '".$id);
}
You don't really need two queries. You should just be able to update like this
mysql_query("UPDATE frases SET `count` = `count` + 1 WHERE id = ".$id);
You didn't close the single quote at the end of the update statement:
mysql_query("UPDATE frases SET count = '".$ntcounter[count]+$add."' WHERE id = '".$id."'")
I need to copy the value in a column named TEAM from one row into another row. Both rows need to have the same team name. This is my query that doesn't work:
$query = "UPDATE profiles SET team = (SELECT team FROM profiles WHERE id = '$coach_id') WHERE id = '$player_id'";
I have tried removing single quotes, removing "FROM profiles", changing value to table.value, tried to give a newdata.clan alias, and I have even tried changing the values to integers instead of parameters. Nothing works, and this is what I get:
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 'WHERE id = '') WHERE id = ''' at
line 3
$query1 = "SELECT team FROM profiles WHERE id = '$coach_id'";
/* get the value of the first query and assign it to a variable like $team_name */
$query2 = "UPDATE profiles SET team = '$team_name' WHERE id = '$player_id'";
Also, you should surround your PHP variables in curly braces:
$query = "UPDATE profiles SET team = \"(SELECT team FROM profiles WHERE id = '{$coach_id}')\" WHERE id = '{$player_id}'";
From the MySQL manual:
"Currently, you cannot update a table
and select from the same table in a
subquery."
Source: http://dev.mysql.com/doc/refman/5.0/en/update.html
Use the method that FinalForm wrote:
<?
$coach_id = 2;
$player_id = 1;
$query1 = "SELECT team FROM profiles WHERE id = '$coach_id'";
$rs = mysql_query($query1);
if ($row = mysql_fetch_array($rs)) {
$team_name = $row['team'];
$query2 = "UPDATE profiles SET team = '$team_name' WHERE id = '$player_id'";
mysql_query($query2);
// Done, updated if there is an id = 1
} else {
// No id with id = 2
}
?>