Populate an array with record ID's that match certain criteria (Select, Where)

Scriptcase 9 General
#1

Hi There,

is there an easy way to create a select where that populates an array (comma separated) with the ID’s of the records that match the criteria?

#2

With group_concat() in your sql select you can group all the records (as defined by your where clause) of a specific field using any separator you want

#3

Hi robdago. thanks for the reply.

that did the trick!

for all those that search the same answer.

just curious, now it is comma sepearted, how can i change this to: for example ;

// Check for record
$check_sql = “SELECT GROUP_CONCAT(wk_id)”
. " FROM ILP_rooster_TEMP"
. " WHERE ev_id = ‘82’"
. " ORDER BY ronde_id";
sc_lookup(rs, $check_sql);

if (isset({rs[0][0]})) // Row found
{
$test = {rs[0][0]};

}
else // No row found
{
$test = ‘’;

}

echo $test;

#4

MySQL | Group_CONCAT() Function - GeeksforGeeks

#5

with replace function you can do it:
SELECT REPLACE(GROUP_CONCAT(wk_id), ‘,’,’;’)

#6

GROUP_CONCAT has three parameters:

  • Distinct: It eliminates the repetition of values from result.
  • Order By: It sort the values of group in specific order and then concatenate them.
  • Separator: By default, the values of group are separated by (, ) operator. In order to change this separator value, Separator clause is used followed by a string literal. It is given as Separator ‘str_value’.

See this link for details:

#7

great, thanks all! this is much help-full. sometimes in Scriptcase things works just a little bit different