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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