CODEWITHSUNDEEP
phparraysvariables
Homer_J
Homer_J

Reputation: 3323

PHP variable results array

I have the following code: 

echo"<input  name=\"q". $i. "\" value=\"4\" id=\"q1a1\" type=\"radio\" ";
    if ($results['q1']==4) echo "checked"
    echo";
    />";

Edit: Added the echo statement, which is why the \ are in situ.

The part $results['q1'] I'd like the 1 of the 'q1' to be a variable - this doesn't seem to have the desired effect:

if ($results['q".$i."']==4) echo "checked"

Upvotes: 0

Views: 56

Answers (4)

Teena Thomas
Teena Thomas

Reputation: 5239

ou can either do 

 if ( $results["q".$i] == 4 ) echo "checked"; // .(dot) operator concatenates strings/variables.

or 

 if ( $results["q$i"] == 4 ) echo "checked"; // double-quotes evaluates variables inside them.

Upvotes: 0

charly
charly

Reputation: 956

you want to write: 

if ($results["q".$i]==4) echo "checked"

the "." operator will concatenate 2 strings to make a string and an associative array takes a string. Moreover variable inside ' (quote) operators won't be parsed (that's the difference between " and ') so '$i' will just give a string $i.

Upvotes: 0

Andrew Leap
Andrew Leap

Reputation: 956

if ($results['q'.$i]==4) echo "checked"

this prevents issues with string interpolation. Wouldn't really be an issue in this case, but I feel better staying away from it, as it makes code look dirtier to me.

Upvotes: 0

JvdBerg
JvdBerg

Reputation: 21856

if you put the string and variable in double quotes, the whole thing is evaluated to 1 variable:

if ( $results["q$i"] == 4 ) echo "checked"

Upvotes: 4

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