How to add or display dynamic array variables

Question

I am converting my multi dimensional post variables to dynamic variables as below:

foreach($_POST as $k => $v){
   ${$k} = $v;
}

So my new array looks like this:

Array(
   [name] => Joe
   [surname] => Blogs
   [study] => Array
        (
            [0] => English
            [1] => IT
        )
   [school] => Array
        (
            [0] => Array
                (
                    [0] => Some School Name
                    [1] => 03/09/2015
                    [2] => Present
                )    
        )
)

So if I want to get the school name, this code will work:

echo $school[0][0];

However, I am struggling to use this variable in an sql statement like below:

$sql = "INSERT INTO table (name, surname, subject_1, subject_2, school1_name, school1_datefrom, school1_dateto) VALUES ('$name', '$surname', '$subject[0]', '$subject[1]', '$school[0][0]', '$school[0][1]', '$school[0][2]', '$school[0][3]')";

echo $sql;

All variables that are not an array or single level array like study are getting displayed fine but the school variables like $school[0][0] are displaying as 'Array[0]', 'Array[1]'......... Why is it doing this and is there away I can get those variables to display correctly?

RiggsFolly · Answer 1 · 2015-09-04T12:18:19.497

2

If you wrap the arrayed values in {} then it should work as you have it. I cannot remember the reasoning behind this but try it.

$sql = "INSERT INTO table (name, surname, subject_1, subject_2, 
                           school1_name, school1_datefrom, 
                           school1_dateto) 
              VALUES ('$name', '$surname', '{$subject[0]}', 
                      '{$subject[1]}', '{$school[0][0]}',
                       '{$school[0][1]', '{$school[0][2]}', 
                       '{$school[0][3]}')";

I remember now is called Complex (curly) syntax

Not because the syntax is complex, but because it allows for the use of complex expressions.

edited Sep 04 '15 at 12:18

answered Sep 04 '15 at 12:10

RiggsFolly

93,638
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thank you, that has worked. i think for simplicity, i will wrap each variable in a curly bracket including the variables that are not an array and just remember this syntax as a rule when i come to add variables in an sql query – Ahm3d Sep 04 '15 at 12:32

score 0 · Answer 2 · edited May 23 '17 at 10:27

0

Wrap the variable in {}, It should work. Curly braces are used to explicitly specify the end of a variable name.

Quickly...

$number = 4;
print "You have the {$number}th edition book";
//output: "You have the 4th edition book";

Wihtout curly braces PHP try to find a variable named $numberth, that not exist!

hope this helps.

Ref.

edited May 23 '17 at 10:27

Community

1
1

answered Sep 04 '15 at 12:15

Navneet

4,543
1
19
29

score 0 · Answer 3 · answered Sep 04 '15 at 12:17

replace

$sql = "INSERT INTO table (name, surname, subject_1, subject_2, school1_name, school1_datefrom, school1_dateto) VALUES ('$name', '$surname', '$subject[0]', '$subject[1]', '$school[0][0]', '$school[0][1]', '$school[0][2]', '$school[0][3]')";

with

$sql = "INSERT INTO table (name, surname, subject_1, subject_2, school1_name, school1_datefrom, school1_dateto) VALUES ('$name', '$surname', '{$subject[0]}', '{$subject[1]}', '{$school[0][0]}', '{$school[0][1]}', '{$school[0][2]}', '{$school[0][3]}')";

How to add or display dynamic array variables

3 Answers3