PHP : Select from Database

Question

thank you for reading. I'm having a little problem with the next sentence:

$nats = mysql_query("SELECT id,name FROM reportSatus WHERE reportId = ". $_POST['id']);
$rnat = mysql_fetch_array($nats);

With print_r($rnat) :

Array(
[0] => 1
[id] => 1
[1] => Poca contaminacion
[name] => Poca contaminacion
[2] => 1
[reportId] => 1)

But in the database with the same sentence is:

id            name
1       Poca contaminacion
2       Mucha contaminacion

Any idea what can it be? Thank you in advance ~

You've fetched a row only once - thus you're getting a one row. If you checked the documentation you would see how it's supposed to use it if you want to fetch multiple rows. — zerkms, Mar 16 '14 at 22:12
`while($row = mysql_fetch_array($nats)){ echo $row['name'] }` — samayo, Mar 16 '14 at 22:14
What's the problem? What do you want to achieve? Because I can see two things that might be confusing for novice PHP/mysqli users here. Besides, this is so dangerous, `". $_POST['id']`. Read about [SQL Injection](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php). — Francisco Presencia, Mar 16 '14 at 22:16

jx12345 · Accepted Answer · 2014-03-16T22:46:00.627

1

try :

echo '<table><tr><th>id</th><th>name</th></tr>';
while ($row = mysql_fetch_assoc($nats)) {
    echo "<tr><td>{$row['id']}</td><td>{$row['name']}</td></tr>";
}
echo '</table>';

edited Mar 16 '14 at 22:46

answered Mar 16 '14 at 22:15

jx12345

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Now it prints : Array ( [0] => 2 [id] => 2 [1] => Mucha contaminacion [name] => Mucha contaminacion ) – Maky-chan Mar 16 '14 at 22:17
ok, try updated code above - is that closer to what you want? – jx12345 Mar 16 '14 at 22:20
tried. No luck. Print this, missing the first one: 2 Mucha contaminacion – Maky-chan Mar 16 '14 at 22:23
try $nats = mysql_query("SELECT id,name FROM reportSatus"); and see if they all come out – jx12345 Mar 16 '14 at 22:25
ok, good. Are you sure that the record "id = 2, name = Mucha contaminacion" actually has a reportId = 1? – jx12345 Mar 16 '14 at 22:30
Yes. Btw, I fixed it somehow. I erased below the query : $rnat = mysql_fetch_array($nats); And it works now... No idea. Thanks for your time. – Maky-chan Mar 16 '14 at 22:34
ah, did you have `$rnat = mysql_fetch_array($nats);` as well as the loop? if so that will have read in the first record and moved the pointer onto the next record before we started printing them out. – jx12345 Mar 16 '14 at 22:43

score 0 · Answer 2 · answered Mar 16 '14 at 22:13

0

From documentation

Returns an array of strings that corresponds to the fetched row, or FALSE if there are no more rows. The type of returned array depends on how result_type is defined. By using MYSQL_BOTH (default), you'll get an array with both associative and number indices. Using MYSQL_ASSOC, you only get associative indices (as mysql_fetch_assoc() works), using MYSQL_NUM, you only get number indices (as mysql_fetch_row() works).

That's actually the way it works, if you need more rows to be returned you will need to loop throw the records

while($rnat = mysql_fetch_array($nats))
{
    //print here
}

answered Mar 16 '14 at 22:13

Fabio

23,183
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Still missing values. Prints: Array ( [0] => 2 [id] => 2 [1] => Mucha contaminacion [name] => Mucha contaminacion ) – Maky-chan Mar 16 '14 at 22:17
It returns what you are selecting, if you loop then it will return all available records for your select – Fabio Mar 16 '14 at 22:19

PHP : Select from Database

2 Answers2