Viewed   54 times

Alright, PHP is throwing this error at me (in the log) when I run the code mentioned below:

Error

mysql_num_rows() expects parameter 1 to be resource, string given in (place) on line 10

Line 9-11

$queryFP = ("SELECT * FROM db");
$countFP = mysql_num_rows($queryFP);
$aID = rand(1, $countFP);

I think it has something to do with the $queryFP's syntax, but I'm not completely sure how to fix it since $queryFP's syntax is the simplest query I've ever seen.

 Answers

3

You need to query the database first.

$queryFP = ("SELECT * FROM db");

Should be:

$queryFP = mysql_query("SELECT * FROM db");
Monday, September 12, 2022
2

I would say just build it yourself. You can set it up like this:

$query = "INSERT INTO x (a,b,c) VALUES ";
foreach ($arr as $item) {
  $query .= "('".$item[0]."','".$item[1]."','".$item[2]."'),";
}
$query = rtrim($query,",");//remove the extra comma
//execute query

Don't forget to escape quotes if it's necessary.

Also, be careful that there's not too much data being sent at once. You may have to execute it in chunks instead of all at once.

Saturday, November 5, 2022
2

You need to give the array_shift() the parameter! Look this example:

$stack = array("orange", "banana", "apple", "raspberry");
$fruit = array_shift($stack); // Here you give the parameter
print_r($fruit);

You give the null parameter on array_shift() and you need to change it!

Update:

array_shift() shifts the first value of the array off and returns it, shortening the array by one element and moving everything down. All numerical array keys will be modified to start counting from zero while literal keys won't be touched. Read here for more

Wednesday, October 19, 2022
 
shark
 
1

Sigh, I'm dumb! At least I learned something today.

The phpinfo() came from the php version of the console, not Apache. Turns out php7.1 and mysqlnd was not loaded in Apache.

Everything is fine now...

Saturday, August 20, 2022
2

The function you're looking for is find_in_set:

 select * from ... where find_in_set($word, pets)

for multi-word queries you'll need to test each word and AND (or OR) the tests:

  where find_in_set($word1, pets) AND find_in_set($word2, pets) etc 
Wednesday, August 17, 2022
Only authorized users can answer the search term. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :
 
Share