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Help with simple php
Posted: Fri May 22, 2009 2:08 pm
by musson
I only want to make a picture apear for each item I will have in my db. whats wrong in that script ? field3-name is a picture url.
Code: Select all
echo "<b>$field1-name
$field2-name2</b><br> <img src="$field3-name"><br>$field4-name<br>$field5-name<hr><br>";
$i++;
}
thanks
Re: Help with simple php
Posted: Fri May 22, 2009 2:19 pm
by anand
musson wrote:I only want to make a picture apear for each item I will have in my db. whats wrong in that script ? field3-name is a picture url.
echo "<b>$field1-name
$field2-name2</b><br> <img src="$field3-name"><br>$field4-name<br>$field5-name<hr><br>";
$i++;
}
thanks
Code: Select all
echo "<b>$field1-name
$field2-name2</b><br> <img src='$field3-name'><br>$field4-name<br>$field5-name<hr><br>";
$i++;
}
You have used " rather you should have used ' .
Try the above code once and see if it works.
Re: Help with simple php
Posted: Tue May 26, 2009 4:37 pm
by musson
I did what you propose me and It still not working. Maybe there is an error in my script:
Code: Select all
mysql_select_db("$database", $con);
$result = mysql_query("SELECT * FROM product ORDER BY ID");
$num=mysql_numrows($result);
echo "<b><center>Database Output</center></b><br><br>";
$i=0;
while ($i < $num) {
$field1-name=mysql_result($result,$i,"name");
$field2-name=mysql_result($result,$i,"price");
$field3-name=mysql_result($result,$i,"picture");
$field4-name=mysql_result($result,$i,"field4-name");
$field5-name=mysql_result($result,$i,"field5-name");
echo "<b>$field1-name
$field2-name2</b><br>
<img src='$field3-name'><br>
$field4-name<br>$field5-name<hr><br>";
$i++;
}
mysql_close($con);
?>
Re: Help with simple php
Posted: Tue May 26, 2009 4:48 pm
by mischievous
Code: Select all
<?php
mysql_select_db("$database", $con);
$result = mysql_query("SELECT * FROM product ORDER BY ID");
$num = mysql_numrows($result);
echo "<h1><div align='center'>Database Output</div></h1>";
if ($num != 0){
foreach($result->result() as $field)
{
echo("<b>".$field->field1." ".$field->field2."</b><br/><img src='".$field->field3."'><br/>".$field->field4."<br/>".$field->field5."<hr/><br/>");
}
} else {
echo("No Results Found");
}
mysql_close($con);
?>
Re: Help with simple php
Posted: Tue May 26, 2009 7:19 pm
by mikemike
I'm sorry if I am wrong as I have not tested but 'mischievous's code appears incorrect to me as he's trying to use an undefined object. Please see my code for the differences:
Code: Select all
# <?php
# mysql_select_db("$database", $con);
# $result = mysql_query("SELECT * FROM product ORDER BY ID");
# $num = mysql_numrows($result);
# echo "<h1><div align='center'>Database Output</div></h1>";
#
# if ($num != 0){
# foreach($result->result() as $field)
# {
# echo("<b>".$field->field1." ".$field->field2."</b><br/><img src='".$field->field3."'><br/>".$field->field4."<br/>".$field->field5."<hr/><br/>");
# }
# } else {
# echo("No Results Found");
# }
# mysql_close($con);
#
# ?>
#
Mine:
Code: Select all
<?php
mysql_select_db($database, $con);
$result = mysql_query("SELECT * FROM product ORDER BY ID");
$num = mysql_num_rows($result);
echo "<h1><div align='center'>Database Output</div></h1>";
if ($num != 0){
while($row = mysql_fetch_assoc($result))
{
echo("<b>".$row['field1']." ".$row['field2']."</b><br/><img src='".$row['field3']."'><br/>".$row['field4']."<br/>".$row['field5']."<hr/><br/>");
}
} else {
echo("No Results Found");
}
mysql_close($con);
?>
php.net states that mysql_query returns a "resource on success, or FALSE on error", a resource is not an object - it is an external resource used by special PHP functions.
And remember to have an alt in your img or it won't validate!
Mike