1
Table example:
| Ti | Orien |
| ---| JML |
| ---| JML |
| ---| RGM |
| ---| AAA |
| ---| AAA |
Result example: JML - 2 RGM - 1 AAA - 2
1
Table example:
| Ti | Orien |
| ---| JML |
| ---| JML |
| ---| RGM |
| ---| AAA |
| ---| AAA |
Result example: JML - 2 RGM - 1 AAA - 2
1
With php5.5 can combine function array_column()
that extract the input values and return them as a new array, with use of the function array_count_values()
that will tell you how many equal values there are in the array, the keys are the database values and the values are the number of occurrences.
$arr = [['str' => 'abc'], ['str' => 2015], ['str' => 2015], ['str' => 'doge'], ['str' => 'wow'], ['str' => 'wow']];
$itens = array_count_values(array_column($arr, 'str'));
echo "<pre>";
print_r($itens);
Return:
Array
(
[abc] => 1
[2015] => 2
[doge] => 1
[wow] => 2
)
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I was thinking something like a Select Ti, Count(Orien) as in a FROM table".
– Rui Manuel Martins
@Ruimanuelmartins , a
select Ti, count(Orien) as num FROM tabela group by Orien
resolves?– rray
I haven’t tested it yet. This is my code:
– Rui Manuel Martins
$query_b="SELECT * FROM
docentes
ORDER BYdocentes
.NomeAscii
ASC"; $query_c= "SELECT Advisor, COUNT() AS numInstances FROM theses GROUP BY Advisor"; print_TD("<select name='orientarisep'></option>"; // list box select command echo "<option value=' '</option>" ." Invite advisor"; foreach ($db->query($query_b) as $Row)//Array or Records stored in $Row { echo "<option value=$Row[Coddocente]>$Row[Coddocente] - $Row[Nomescii]</option>"; / Option values are Added by looping through the array */ } echo "</select>";– Rui Manuel Martins
I’m sorry I was wrong. the goal is basically to dropdown a list with the information of a table and then go to the other table for the records that already have more than one.
– Rui Manuel Martins