php + mysql rows only returning int

Question

Hi I'm starting out on php and I have the following script

<?php

$username = "firstleg_floa";
$password = "**";
$hostname = "***1";
$db = "firstleg_bank";
$guildUser = strtolower('grissiom');

$dbh = mysql_connect( $hostname, $username, $password) or die ("Unable To Connect");
$connectDB = mysql_select_db($db, $dbh);
$results = mysql_query("SELECT * FROM Bank where userId ='" .$guildUser."'");

$i = 0;
$rsArr = array();

While ($row = mysql_fetch_array($results, MYSQL_ASSOC))
{
    $rsArr [$i] [0] = $row{'userId'};
    $rsArr [$i] [1] = $row{'item'};
    $rsArr [$i] [2] = $row{'amount'};
    $rsArr [$i] [3] = $row{'position'};
    $i++;
}
?>
<?="ghdfdgdfg ". $rsArr[$i][1];}." ----";?>
    <? echo $rsArr[$i][0]; ?>
    <table>
    <tr><td>Item</td><td>Amount</td></tr>
    <?for ($x=0;$x <= $i; $x++)
    {?>
    <tr><td><?=$rsArr[$x][3];?></td><td><?=$rsArr[$x][2];?></td></tr>
    <?}?>
    </table>

<?php
mysql_close($dbh); ?>

and I have the following data in the database

1 - grissiom - Silk Cloth - 100 - 1

with the above data and the above script all I can manage to pull out is the amount(100) and the position(1) how come i am unable to pull anything else out? can someone help me please

Jeremy DeGroot · Accepted Answer · 2009-03-26 15:38:27Z

4

It could be because you're using curly braces instead of square. The row returned by mysql_fetch_array() is a PHP array, not an object or string.

answered Mar 26, 2009 at 15:38

Jeremy DeGroot

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3 Comments

Ivan Over a year ago

arrays in php may also be defined with curly braces. this is not a problem.

Ivan Over a year ago

Milen, try running this code: <?php $arr = array('test' => 'a test','another test2' => 'test',); echo $arr['test']."\n"; $arr{'test3'} = 'third element'; var_export($arr); echo "\n"; $arr2 = array('test1','test2','test3'); echo $arr{2}."\n"; array_push($arr2,'test4'); var_export($arr2); ?>

Ivan Over a year ago

the only problem is that you can't define $arr{} = 'value'

karim79 · Accepted Answer · 2009-03-26 23:46:20Z

3

Possible solutions:

1) Those column names in the DB have different case, e.g. userId could be just userid

2) Those columns don't exist.

3) You are using curly braces, instead of square brackets:

 $rsArr [$i][0] = $row{'userId'}; //wrong
 $rsArr [$i] [0] = $row['userId']; //correct

Try replacing the while loop with this, it will read out the contents of each record as the DB has passed it to PHP:

while ($row = mysql_fetch_array($results, MYSQL_ASSOC))
{
    echo '<pre>' . var_dump($row) . </pre>;
}

to see exactly what is coming back from the DB.

edited Mar 26, 2009 at 23:46

answered Mar 26, 2009 at 15:41

karim79

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Comments

Ivan · Accepted Answer · 2009-03-26 15:40:46Z

1

after loop is finished $i is equal to 1, but $rsArr[$i] is not defined, so variables $rsArr[$i][0] and $rsArr[$i][1] doensn't exists, try using $rsArr[0][0] and $rsArr[0][1] like this:

<?="ghdfdgdfg ". $rsArr[0][1];}." ----";?>
    <? echo $rsArr[0][0]; ?>
    <table>
    <tr><td>Item</td><td>Amount</td></tr>
    <?for ($x=0;$x <= $i; $x++)
    {?>
    <tr><td><?=$rsArr[$x][3];?></td><td><?=$rsArr[$x][2];?></td></tr>
    <?}?>
    </table>

answered Mar 26, 2009 at 15:40

Ivan

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Comments

Alekc · Accepted Answer · 2009-03-26 15:40:28Z

It seem ok to me, maybe issue is given from using curly braces.

Anyway i'd go with some simpler syntax, i.e:

<?php

$username = "firstleg_floa";
$password = "**";
$hostname = "***1";
$db = "firstleg_bank";
$guildUser = strtolower('grissiom');

$dbh = mysql_connect( $hostname, $username, $password) or die ("Unable To Connect");
$connectDB = mysql_select_db($db, $dbh);
$results = mysql_query("SELECT * FROM Bank where userId ='" .$guildUser."'");

$i = 0;
$rsArr = array();
while ($row = mysql_fetch_array($results, MYSQL_ASSOC))
{
    $rsArr[] = $row;
}
mysql_close($dbh);
echo '<table><tr><td>Item</td><td>Amount</td></tr>';
for ($i = 0;$i<count($rsArr);$i++){
    echo <<<EOF
    <tr><td>{$rsArr["item"]}</td><td>{$rsArr["amount"]}</td></tr>
EOF;
}
echo '</table>';
 ?>

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