So I have a code I want to use
mysql_select_db("website", $con);
$result=mysql_query("SELECT * FROM characters where online='1'");
while ($row=
mysql_fetch_array($result))
{
echo $row['name'];
}
if ($row['race'] == "1");
{
echo '<img src="img/8-0.gif" />';
}
if ($row['class'] == "3");
{
echo '<img src="img/3.gif" ?>';
}
mysql_close($con);
?>
Now I only wanted the two images to show if the online field was 1, but they show no matter what. Does anyone know how I can fix this? Thanks.
For the sake of the argument:
mysql_select_db("website", $con);
$result=mysql_query("SELECT * FROM characters where online='1'");
while ($row= mysql_fetch_array($result))
{
echo $row['name'];
//} This bracket would immediately close your query processing and only display the last images. Or is that the desired behaviour?
if($row['isOnline'] == '1') { //Makes sure, that 'isOnline' is set before displaying.
if ($row['race'] == "1");
{
echo '<img src="img/8-0.gif" />';
}
if ($row['class'] == "3");
{
echo '<img src="img/3.gif" ?>';
}
}
} //This bracket closes the actual query result handling
mysql_close($con);
?>
if ($row['race'] == "1" AND $row['online'] == 1);
{
echo '<img src="img/8-0.gif" />';
}
if ($row['class'] == "3" AND $row['online'] == 1);
{
echo '<img src="img/3.gif" ?>';
}
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