PHP - basic select query returning only the first row of mysql database

Question

I've been running the query oh phpMyAdmin and it shows all the rows, but my query in php only returns the first row.

$result = $mydb -> query("SELECT * FROM music_sheet WHERE category='".$_REQUEST["submitter"]."'"); 

print(count($result)); //always returns 1

for ($x = 0; $x < count($result); $x++) {
    $row = mysqli_fetch_row($result);
}

Answer 1

For reference, here's why count() is returning 1 . From the manual:

If the parameter is not an array or not an object with implemented Countable interface, 1 will be returned. There is one exception, if array_or_countable [ the parameter ] is NULL, 0 will be returned.

Since $result is a resource (object) that is returned from your query, not an array which you would get within your loop getting the actual data out of the resource, count($resource) will return 1 .

Your solution is of course to use mysqli_num_rows() . To retrieve data from this resource, you need to loop over it as you are doing, but either use mysqli_num_rows() in place of count() , or other (more common) ways of looping through a result set, like so:

while($row = mysqli_fetch_row($result)) {
    // do stuff
}

Answer 2

您必须使用mysqli_num_rows($result)函数对MySQLi查询返回的行进行计数。

Answer 3

尝试这个

echo mysqli_num_rows($result);

Answer 4

while($row = mysqli_fetch_row($result)) {
// write your code...
}

Use this instead of the for loop

Answer 5

the first thing that the query method returns to you is a resource/object. This query method always return a mysqli_result object for successful queries using SELECT, SHOW, DESCRIBE or EXPLAIN queries . For other successful queries mysqli_query() will return TRUE. For that reason it always count it as "1", what you should try is:

$result = $mydb -> query("SELECT * FROM music_sheet WHERE category='".$_REQUEST["submitter"]."'"); 

$numRows = $result->num_rows;

print(count($numRows)); //it should show you the total amount of rows

//verify the amount of rows before of tryng to loop over it
if ($numRows) {
    while($object = mysqli_fetch_object($result)){
    //for each loop you will have an object with the attributes of the row
    echo $object->song_name; 
    }
}

This should work,

Regards.