I have looked all over the web and cannot seem to find a proper way to assign the result of a MySQL query to a PHP variable. The code that I currently have returns a "resource id #3" error. Here is what it looks like.
//Select the 'aQID' of the question that has it's BOOL set to "true"
$currentQ = mysql_query("SELECT aQID FROM approvedQuestions WHERE status='1'");
$cuQ = mysql_result($currentQ,1,"status");
echo $cuQ;
I know that the query will only ever return one record (the active question). But I cannot seem to figure out what function to use.
$currentQ = mysql_query("SELECT aQID FROM approvedQuestions WHERE status='1'");
$rs = mysql_fetch_array($currentQ);
var_dump($rs);
Related
My query is not working when I use the variable in the WHERE clause. I have tried everything. I echo the variable $res, it shows me the perfect value, when I use the variable in the query the query is not fetching anything thus mysqli_num_rows is giving me the zero value, but when I give the value that the variable contains statically the query executes perfectly. I have used the same kind of code many times and it worked perfectly, but now in this part of module it is not working.
Code:
$res = $_GET['res']; // I have tried both post and get
echo $res; //here it echos the value = mahanta
$query = "SELECT * FROM `seller` WHERE `restaurant` = '$res'"; // Here it contains the problem I have tried everything. Note: restaurant name is same as it is in the database $res contains a value and also when I give the value of $res i.e. mahanta in the query it is then working.
$z = mysqli_query($conn, $query);
$row2 = mysqli_fetch_array($z);
echo var_dump($row2); // It is giving me null
$num = mysqli_num_rows($z); // Gives zero
if ($num > 0) {
while ($row2 = mysqli_fetch_array($z)) {
$no = $row2['orders'];
$id = $res . $no;
}
}
else {
echo "none selected";
}
As discussed in the comment. By printing the query var_dump($query), you will get the exact syntax that you are sending to your database to query.
Debugging Tip: You can also test by pasting the var_dump($query) value in your database and you will see the results if your query is okay.
So update your query syntax and print the query will help you.
$query = "SELECT * FROM `seller` WHERE `restaurant` = '$res'";
var_dump($query);
Hope this will help you and for newbies in future, how to test your queries.
Suggestion: Also see how to write a mysql query syntax for better understanding php variables inside mysql query
The problem is the way you're using $res in your query. Use .$res instead. In PHP (native or framework), injecting variables into queries need a proper syntax.
I need to be able to check and see in a certain string is anywhere within my SQL table. The table I am using only has one column of char's. Right now it is saying that everything entered is already within the table, even when it actually is not.
Within SQL I am getting the rows that have the word using this:
SELECT * FROM ADDRESSES WHERE STREET LIKE '%streeetName%';
However, in PHP the word is being entered by the user, and then I am storing it as a variable, and then trying to figure out a way to see if that variable is somewhere within the table.
$duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
if(!empty($duplicate))
{
echo "Sorry, only one of each address allowed.<br /><hr>";
}
You need to do a little bit more than building the query, as mysql_query only returns the resource, which doesn't give you any information about the actual result. Using something like mysql_num_rows should work.
$duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
if(mysql_num_rows($duplicate))
{
echo "Sorry, only one comment per person.<br /><hr>";
}
Note: the mysql_* functions are deprecated and even removed in PHP 7. You should use PDO instead.
In the SQL you used
%streeetName%
But in the query string below, you used
%$streeetName%
Change the correct one
$duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
if(!empty($duplicate))
{
echo "Sorry, only one comment per person.<br /><hr>";
}
if($results->num_rows) is what you need to check if you have results back from your query. An example of connection and query, check, then print or error handle, the code is loose and not checked for errors. Best of luck...
//Typically your db connect will come from an includes and/or class User...
$db = new mysqli('localhost','user','pass','database');
$sql = "SELECT * FROM `addresses` WHERE `street_name` LIKE '%$streetName%'",$connect;
//test your queries in PHPMyAdmin SQL to make sure they are properly configured.
//store the results of your query in a variable
$results = $db->query($sql);
$stmt = '';//empty variable to hold the values of the query as it runs through the while loop
###########################################################
#check to see if you received results back from your query#
###########################################################
if($results->num_rows){
//loop through your results and echo or assign the values as needed
while($row = $results->fetch_assoc()){
echo "Street Name: ".$row['STREET_NAME'];
//define more variables from your DB query using the $row[] array.
//concatenate values to a variable for printing in your choice further down the document.
$address .= $row['STREET_NAME'].' '.$row['CITY'].' '$row['STATE'].' '$row['ZIP'];
}
}else{ ERROR HANDLING }
I have this function:
function set_spell() {
global $connection;
$query = "SELECT * FROM Spells";
$result_set = mysqli_query($connection, $query);
$spell_id_list = mysqli_fetch_array($result_set);
return $spell_id_list;
Then I try to output a random ID from the array, but I always get ID 1.
$spell_id_list = set_spell();
echo $spell_id_list[array_rand($spell_id_list)];
When I run this query in MYSQL, I get a list of all ID's as expected. Why doesn't the code above select one at random?
This is probably a stupid questions that I'll smack myself after I see the answer... but I've been stuck for longer than I think I should be on it.
Thanks for the help.
You just returning only single result from your function so either create array of all fetched results & return the same, or simply use order by rand() in your query,
SELECT * FROM Spells ORDER BY RAND();
Hey all i am trying to figure out a way to see what all values are within the array:
$sql = "select * from product where type = ? limit 1";
$query = self::$__CONN__->prepare($sql);
$query->execute(array($type)) or die(print_r($query->errorinfo()));
I am new at PHP and im not sure what all that is doing to populate the query string. I am guessing that the first query is connecting to the server and sending out the query but how do i see what its returning back in the array above?
Any help would be great!
Just dump the $type variable
var_dump($type);
Or if you are interested in the array wrapper itself
var_dump(array($type));
I'm new to PHP and SQL, but I need a way to store the result of an SQL Query into a variable.
The query is like this:
$q = "SELECT type FROM users WHERE username='foo user'";
$result = pg_query($q);
The query will only return one string; the user's account type, and I just need to store that in a variable so I can check to see if the user has permission to view a page.
I know I could probably just do this query:
"SELECT * FROM users WHERE username='foo user' and type='admin'";
if(pg_num_rows($result) == 1) {
//...
}
But it seems like a bad practice to me.
Either way, it would be good to know how to store it as a variable for future reference.
You can pass the result to pg_fetch_assoc() and then store the value, or did you want to get the value without the extra step?
$result = pg_query($q);
$row = pg_fetch_assoc($result);
$account_type = $row['type'];
Is that what you are looking for?
Use pg_fetch_result:
$result = pg_query($q);
$account_type = pg_fetch_result($result, 0, 0);
But on the other hand it's always good idea to check if you got any results so I'll keep the pg_num_rows check.