为什么我没有在查询代码中得到所有结果？

Question

I am having a problem. The following code works fine in my local, but on the live server, it's not working properly..I was supposed to get two rows, but on the live server I am getting only 1 result.

$query = mysql_query('SELECT * FROM `wspm_t_colors`');
$result = mysql_fetch_object($query);
print json_encode($result);

What could possibly be the error ?...

Answer 1

I can't believe you get two rows with this, to get all rows you have to do like this:

$query = mysql_query('SELECT * FROM `wspm_t_colors`');
while($result = mysql_fetch_object($query))
{
    print json_encode($result);
}

mysql_fetch_object/array/row always returns only one row and moves the pointer to the next row, if there is no next row it returns false.

Answer 2

I think this is because mysql_fetch_object only returns a single row result. You need something like this:

$query = mysql_query('SELECT * FROM `wspm_t_colors`');
while($row = mysql_fetch_array($query))
{
   \\access each result here
}

Answer 3

Your code is only getting one row. The mysql_fetch_object() function only returns one row. You need to try something like this:

$query = mysql_query('SELECT * FROM `wspm_t_colors`');
$json = array();
while ($result = mysql_fetch_object($query))
    $json[] = $result;
print json_encode($json);

Answer 4

I can't see how you can have two rows in local

$array = array();
$query = mysql_query('SELECT * FROM `wspm_t_colors`');
while($result = mysql_fetch_object($query))
{
    $array[] = $result;
}
print json_encode($array);