CODEWITHSUNDEEP
phpmysqlmultiple-conditions
Bogdan Constantinescu
Bogdan Constantinescu

Reputation: 13

While, if and else conditions

I have the code:

while($res = mysql_fetch_array($rst)) 
        if(($res)==TRUE)
         echo "$res[2]";
        else
         echo "<img src='imagini/banner.png'>";

First echo it shows me output when I have results. Second echo doesn`t show the image when if condition is false. Any help, please? :) Thanks.

Upvotes: 0

Views: 225

Answers (3)

ArtisticPhoenix
ArtisticPhoenix

Reputation: 21661

A few formatting things I would fix that increase the readability

while($res = mysql_fetch_array($rst)) 
    if(($res)==TRUE)
     echo "$res[2]";
    else
     echo "<img src='imagini/banner.png'>";

To this

while(false !== ( $res = mysql_fetch_array($rst) ) ){
    if($res[2] == true){ // or isset( $res[2] ) ?
         echo "$res[2]";
    }else{
         echo "<img src='imagini/banner.png'>";
    }
}

Lets start at the top. mysql_fetch_array http://php.net/manual/en/function.mysql-fetch-array.php ~ returns an array or false.

Then we are evaluating it against false, and assigning it's value to $res. Next the extra () around $res is meaningless, even checking true here Likely your issue is meaning less, it will never enter the loop as false, so inside the loop $res is never false. Last addition of proper indenting and {brackets}.

In reality you probably intend this?

if( mysql_num_rows( $rst ) ){
    while(false !== ( $res = mysql_fetch_array($rst) ) ){        
        echo $res[2];    
    }
}else{
    echo '<img src="imagini/banner.png">';
}

In the equivalent PDO 

if( $stmt->rowCount() ){
    while(false !== ( $res = $stmt->fetch(PDO::FETCH_NUM) ) ){        
        echo $res[2];    
    }
}else{
    echo '<img src="imagini/banner.png">';
}

Pleas note mysql_* family of functions are depreciated as of PHP5.5 http://php.net/manual/en/book.pdo.php

Upvotes: 1

cobarzan
cobarzan

Reputation: 682

Most probably the relative path is at fault here. Make sure that is correctly set.

EDIT

While the path thing might be a problem, it is not the biggest here.

What you most probably want is either:

// this tests if there are zero or more rows in your result
if (mysql_num_rows($rst)==0)
{
    echo "<img src='imagini/banner.png'/>";
}
else
{
    while ($res=mysql_fetch_array($rst))
    {
         echo $res[2];
    }
}

or

// this tests, for each row, if a column is set or not
while ($res=mysql_fetch_array($rst))
{
    // index is the index of the column that can be set or not
    if (empty($res[index]))
    {
        echo "<img src='imagini/banner.png'/>";
    }
    else
    {
         echo $res[2];
    }
}

Also, you seem to retrieve <img> tags from the database, since you have an default <img> tag as an alternative. If $res[2] has this pattern <img src='some_path'/> maybe you can modify the content of the database to keep just the relative path (without the tag) and echo the tag in the while loop.

Upvotes: 1

Aaron G.
Aaron G.

Reputation: 11

Try switching the if and else so you can see if it has something to do with what your outputting or if it is failing to reach the else. 

    if(($res)==FALSE)
     echo "<img src='imagini/banner.png'>";

Upvotes: 0

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