How to extract mysql data into json using php?

[semanticnotion]

i have retrieved mysql data from one table in json using the following script

$table_first = 'abc';
$query = "SELECT * FROM $table_first";
$resouter = mysql_query($query, $conn);
 
$set = array();
 
$total_records = mysql_numrows($resouter);
if($total_records >= 1){
 
  while ($link = mysql_fetch_array($resouter, MYSQL_ASSOC)){
    $set[] = $link;
  }
}
 
echo json_encode($set);

how can i retrieved data from two other tables in which there is a foreign key of this table in both of those tables. OR simply how can i retrieved data from 3 mysql tables in php.

[code green]

I don't know what json does but the query will need two joins.
Maybe it is a question better asked on the MySQL forum.
I doubt you will want SELECT * when joining three tables

[semanticnotion]

the join query returns all the data at once what i want to do is that to retrieved the relevant records. i.e for id 1 of first table there are 5 relevant records in second table and 3 records relevant to this in third table.
i want to print the id 1 of table one than all the relevant records of table 2 and table 3.
And same for id 2 of table 1......

[code green]

the join query returns all the data at once what i want to do is that to retrieved the relevant records

Which is why I suggested not to use SELECT *.
Your SELECT string should detail which fields you require.
I can't help anymore without the table structures and the required output in tabular form

[semanticnotion]

my table structure is

recipe{rec_id,name,overview,category,time,image}
ingredients{ingredient_id,ingredient_name,ammount, rec_id}
instructions{instruction_id,instruction_text,rec_i d}

look at rec_id which is foreign key in both tables.

so queries should be

$query1="select * from recipe";
 

than it should iterate in table2 against the relevant id from table1;
e.g

$query2="select specific_field from ingredients where rec_id='rec_id from table1';

[code green]

I suggest putting the query question on MySQL forum.
But MySQL won't do what you are expecting it to do.
You will need an explicit JOIN, primary-foreign keys are ignored.