Select query for getting a single value is not working

Question

 $sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
 $result = mysql_query($sql);
 $value = mysql_fetch_object($result);
 $teacheremail2 = $value->temail;
 echo $teacheremail2;

echo $teacheremail2 returns nothing.

$teachername is valid and i have checked multiple times.

Not an answer, but using `LIMIT` without `ORDER BY` is fairly meaningless, because there is no formal logic/reasoning regarding _which_ single record you want to get back. Add an `ORDER BY` clause so that the query makes sense. — Tim Biegeleisen, Aug 29 '18 at 01:59
What is your php version, and do you have display errors enabled? — Don't Panic, Aug 29 '18 at 02:12
https://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display — Don't Panic, Aug 29 '18 at 02:15
Error reporting for PHP http://php.net/manual/en/function.error-reporting.php and `mysql_error()` for the query are two error checking tools for you to use. You can then update your question to contain what those errors were, if any. — Funk Forty Niner, Aug 29 '18 at 02:16
I think that's likely the phpmyadmin version, not the php version — Don't Panic, Aug 29 '18 at 02:17

score 0 · Answer 1 · answered Aug 29 '18 at 02:07

0

It should be a two-dimensional array , you need $value[0]->temail

answered Aug 29 '18 at 02:07

Mr.Zhang

1
1

Pretty sure mysql_fetch_object returns an object. – Don't Panic Aug 29 '18 at 02:10
Sorry no results – Genius Partner Aug 29 '18 at 02:11

arda sugriwo · Answer 2 · 2018-08-29T07:47:27.807

0

The result of mysql_fetch_object($result) is an object(stdClass).

The explanation of object(stdClass) ican be found at this link

$sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
$result = mysql_query($sql);
while ($value = mysql_fetch_object($result))
    {
    $teacheremail2 = $value->temail;
    echo $teacheremail2;
    }

edited Aug 29 '18 at 07:47

answered Aug 29 '18 at 02:55

arda sugriwo

1
2

how mickmackusa ? – arda sugriwo Aug 29 '18 at 06:50

score -1 · Answer 3 · answered Aug 29 '18 at 02:04

-1

First off, you'll want to run the query directly against your database to ensure that the query returns some kind of result. Secondly, if that works, you'll want to echo $value directly to check that you are getting results back on the webpage.

Then you can check if temail is a field of $value

answered Aug 29 '18 at 02:04

Jack

472
6
13

I ran SELECT temail FROM `teacherusers` WHERE `tfullname` = "XXX XXX" and i got the result in the phpmyadmin – Genius Partner Aug 29 '18 at 02:09
Good. So move on to the second step then – Jack Aug 29 '18 at 02:11
echo $value; shows nothing . temail is the column name of the table , double checked – Genius Partner Aug 29 '18 at 02:14
Try using var_dump http://php.net/manual/en/function.var-dump.php Also, ensure that you have set an intial value for $teachername – Jack Aug 29 '18 at 03:32

arda sugriwo · Answer 4 · 2018-08-29T02:45:42.987

-2

$sql = "SELECT temail FROM teacherusers WHERE tfullname='$teachername' limit 1";
 $result = mysql_query($sql);
while ($value = mysql_fetch_object($result))
    {
    $teacheremail2 = $value->temail;
    echo $teacheremail2;
    }

hope this help

edited Aug 29 '18 at 02:45

answered Aug 29 '18 at 02:01

arda sugriwo

1
2

3

This should work either way because of variable interpolation https://stackoverflow.com/questions/5605965/php-concatenate-or-directly-insert-variables-in-string – Jack Aug 29 '18 at 02:03
can you provide me the table and its data? i'll test in my local – arda sugriwo Aug 29 '18 at 02:14

Select query for getting a single value is not working

4 Answers4