PHP "if" not working

Question

I have a Table where i want to Display time someone has worked. This comes from a mySql a Field should Display the minutes, and if a other field ist true, then it should be add "high" after the minutes

I tried this:

Print "<td width=50px>";
PRINT $info['minutes'] if($info['high'] == 0 ) {echo "Hoch"} ;

and got this:

Parse error: syntax error, unexpected 'if' (T_IF) in C:\xampp\htdocs\callinfo.php on line 63

Someone can Help? :)

Missing semi-colon, case issues...

Shomz
– Shomz

2013-11-23 17:55:12 +00:00
Commented Nov 23, 2013 at 17:55 — Shomz
– Shomz, Commented Nov 23, 2013 at 17:55

Duniyadnd · Accepted Answer · 2013-11-23 17:57:02Z

2

You need to space out your code if you can't handle parse errors. Also, print/echo are pretty much same, use one or the other, not both, make your life easier in the future when you're looking for specific code.

PRINT $info['minutes'];
if ( $info['high'] == 0 ) {
    echo "Hoch";
}

answered Nov 23, 2013 at 17:57

Duniyadnd

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Comments

Patato · Accepted Answer · 2013-11-23 17:55:20Z

1

you need to add ;

PRINT $info['minutes'] ;
if($info['high'] == 0 )
{
   echo "Hoch";
}

answered Nov 23, 2013 at 17:55

Patato

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Comments

user3025149 · Accepted Answer · 2013-11-23 17:57:35Z

1

It lacks a ; after PRINT $info['minutes']

PRINT $info['minutes'];

answered Nov 23, 2013 at 17:57

user3025149

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