Passing the values of mysql_fetch_array to an Array()? - php

Question

this is my code:
$scontain = "SELECT id FROM voting";
$qcontain = mysql_query($scontain);
$r_idarray = array();
while ($rcontain = mysql_fetch_array($qcontain))
{
$r_idarray[] = $rcontain['id'];
//let's say there are 5 names here//
}
echo $r_idarray;
I was trying to get the whole content of my table 'voting'. Inside the while loop it successfully prints out the whole content of the table which is under the column 'id' but when I tried to echo it outside the while loop it prints 'Array'? Can anyone knows how to solve this. Thank you in advance...

Answer 1

To print the array you can do:
print_r($r_idarray);
instead of
echo $r_idarray;

Answer 2

You have to use print_r($r_idarray) .
echo is used for printing strings.
use print_r for printing arrays.

Answer 3

Use print_r or var_dump for printing the array
Replace:
echo $r_idarray;
With:
print_r($r_idarray);
Or:
var_dump($r_idarray);

Answer 4

If you want to display whole content
SELECT id FROM voting
To
SELECT * FROM voting
And after that you can use print_r($r_idarray);

Answer 5

You can't "echo" an array, you can only "echo" a string. Of course you will see 'Array'.
In order to print out the contents of an array, you either have to iterate through it's contents, or do a "var_dump($array)".
while($val in $array){
echo $val;
}
or:
print_r($array);
or:
var_dump($array);

Answer 6

One thing youe query should be like this to select all columns
$scontain = "SELECT * FROM voting";
You only have to do this
$i=0;
while ($rcontain = mysql_fetch_array($qcontain))
{
$r_idarray[$i] = $rcontain;
$i++;
}
echo '<pre>';
print_r($r_idarray);
And if you need particular columns from result
$i=0;
while ($rcontain = mysql_fetch_array($qcontain))
{
$r_idarray[$i]['id'] = $rcontain['id'];
$r_idarray[$i]['column1'] = $rcontain['column1'];
$r_idarray[$i]['column2'] = $rcontain['column2'];
$r_idarray[$i]['column3'] = $rcontain['column3'];
$i++;
}
echo '<pre>';
print_r($r_idarray);