How to return entire mysql table with php?

Question

$result = mysql_query("SELECT * FROM users");
$values = array();
while($row = mysql_fetch_array($result)) 
{
$values[] = array($row['tried']);
}
return $values;

That only returns the word array when being called as a webservice. What am I missing or doing wrong?

You are not echoing anything here, so this is not the code where the problem is. You probably need to show the full code — Pekka
– Pekka, Commented Oct 7, 2010 at 23:18
Decide what format you want it in (HTML, XML, json, serialized, comma separated, etc.),and use that. — Wrikken
– Wrikken, Commented Oct 7, 2010 at 23:19
Please explain what kind of web services you're using. Are you using a library? What is the MIME type? — Matthew Flaschen
– Matthew Flaschen, Commented Oct 7, 2010 at 23:20
what is the structure of your data? what do you expect from the column 'tried'? — KalenGi
– KalenGi, Commented Oct 7, 2010 at 23:28

kevtrout · Accepted Answer · 2010-10-07 23:21:50Z

2

It's not returning the word 'array', it's telling you that what is being returned is an array.

To see what is in the array, use either var_dump($values); or print_r($values);. The only difference is the format of the display.

In addition, I don't believe you need to declare array in your assignment to $values.

$values[] = $row['tried'];

should work. $values will still be an array.

answered Oct 7, 2010 at 23:21

kevtrout

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1 Comment

Jon Over a year ago

thanks for this worked perfectly...now just need to figure out this " content type of 'text/html', but expected 'text/xml'" error when it returns $values

Michał Tatarynowicz · Accepted Answer · 2010-10-07 23:41:53Z

0

I guess you can try changing the last line to: return json_encode($values);. It should at least let you see what you're doing (and may even be usable, depending on situation).

answered Oct 7, 2010 at 23:41

Michał Tatarynowicz

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Comments

Explosion Pills · Accepted Answer · 2010-10-08 03:58:03Z

0

You don't have to SELECT * if you are only using one column. Just SELECT that column.

If it is printing out array, you are probably calling

echo $values;

This is not going to work. There are many solutions to see what you want, e.g.

foreach ($values as $val) { echo $val; }

answered Oct 8, 2010 at 3:58

Explosion Pills

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