If then Else help please

[johnber]

Hi

 

Im quite new to php but have managed to create a page which reads records from a mysql database and present them in a table

 

one of the fields in each record is an image filename and I can get it to display the image - not every record has an image and so currently it will just show a broken image so I assume and if then else statement will fix this as if there is no image i can say No Image Available

 

my code looks like this

 

echo "<tr>

<td>{$row['event_id']}</td>

<td>{$row['user_ID']}</td>

<td>{$row['Lat']}</td>

<td>{$row['long']}</td>

<td>{$row['source']}</td>

<td>{$row['Time']}</td>

<td>{$row['Duration']}</td>

<td>{$row['snratio']}</td>

<td>{$row['adjfreq']}</td>

<td>{$row['doppler_Hz']}</td>

<td><a href='/mimages/{$row['image']}' target='_blank'><img src='/mimages/{$row['image']}' width='100' height='100'></a></td>";

echo "</tr>";

 

so its all around the last last line - pointers really appreciated

 

John

 

 

[requinix]

You can't put an if/else inside a string so that means you have to end the one you've built up early.

echo "<tr>
<td>{$row['event_id']}</td>
<td>{$row['user_ID']}</td>
<td>{$row['Lat']}</td>
<td>{$row['long']}</td>
<td>{$row['source']}</td>
<td>{$row['Time']}</td>
<td>{$row['Duration']}</td>
<td>{$row['snratio']}</td>
<td>{$row['adjfreq']}</td>
<td>{$row['doppler_Hz']}</td>
<td>";

Then add in an if/else to output the image if one is present. Tip: you don't need an else.

 

Then finish the row.

echo "</td></tr>";

[johnber]

im close after getting a few syntax issues I now have this

 

<td>{$row['doppler_Hz']}</td>";

 

echo "<td>";

if (is_null.$row['image'])

         echo "No data!";

     else

         echo "<td>{$row['image']}</td>";

echo "</td></tr>";

 

However it shows No data even if there is data, so this statement

 

 if (is_null.$row['image']) is wrong and its probably my understanding of null \ empty ?

 

John B

 

 

[requinix]

is_null is a function like any other so you need to use it as such.

 

And apparently you don't have your environment set up properly for development. Find your php.ini, find and change two settings

display_errors = on
error_reporting = -1

then restart Apache/nginx. Try your code as it is now without the is_null fix, see what PHP says, then make the change.

[johnber]

learning a lot so error reporting now on and after a few reported syntax errors which I fixed I have

echo "<td>";
if ($row['image'])
    echo "{$row['image']}</td>";
else
    echo "No data!";
echo "</td></tr>";

 

which works a treat - thanks for all the help. John

[ginerjm]

I know people love to use the nuances of PHP's true/false capabilities.  But - especially since you are new - try and write Meaningful Code instead of these things you have to think about when seeing them.

 

 

if ($row['image'] == '')

 

is much more direct than the line you copied from somewhere.  Or even use the is_null function that you mis-used earlier.  Lines like that are so much clearer when reviewing lots of code and not having to stop and think about them.