num_rows和count（）不匹配-PHP

Question

$result = $conn->query($sql);
echo $result->num_rows; //2 is value produced

BUT

$tagResult =  mysqli_fetch_array($result,MYSQLI_NUM);
echo count($tagResult); //1 is the value produced

Isn't it suppose to produce the same number of values?

Answer 1

mysqli_fetch_array()

Returns an array of strings that corresponds to the fetched row or NULL if there are no more rows in resultset.

So it only fetches 1 row. To get both/all you need to fetch all rows in a loop:

while($tagResult[] =  mysqli_fetch_array($result,MYSQLI_NUM)) {}
echo count($tagResult);

Answer 2

mysqli_fetch_array will fetch one row at a time, wherever the pointer for the result currently is. When you first get the result, the pointer is at the first row. When you call mysqli_fetch_array, it will grab record 1 (where the pointer is) then increment the pointer to the next record. When you call it again, it will grab record 2, and so on.

If you have more than one row returned, you will need to loop through the results, calling mysqli_fetch_array on each row. Because mysqli_fetch_array returns null when there are no more rows, you can use a simple while loop:

$result = $conn->query($sql);
echo $result->num_rows; //2 is value produced

$count = 0;
while ($tagResult = mysqli_fetch_array($result,MYSQLI_NUM))
{
    print_r($tagResult); // Output each row
    $count ++;
}
echo $count; //outputs 2