Error passing variable as double parameter

Question

Error passing variable as double parameter

Navigation

#1 by (4 votes)
#2 by (3 votes)

3

I have a function

pesquisaPagamentos($pesquisa)

I have a $pesquisa variable that is getting the following value: '2015-10-05','2015-10-01' , with single quotation marks.

It turns out that the way it, when it arrives at function

<?php 
  if(isset($_GET["acao"]) && $_GET["acao"] == "listaArr") {

      $pesquisa = (isset($_POST["dataIni"])) ? "'".$PhpUtil->formataData($_POST["dataFim"])."','".$PhpUtil->formataData($_POST["dataIni"])."'" : "'',''";



print "<pre>";                                  
print_r($pesquisa);                                 
print "</pre>";                                 

      $arrecadacaoDia = $rel->pesquisaPagamentos($pesquisa);

I try to print the value of the first parameter and it comes with the integer value of the $ search variable, that is: '2015-10-05','2015-10-01' . The second parameter of the function receives an empty value.

php

asked by anonymous 05.10.2015 / 23:53

2 answers

Using case in field null Is there a pattern similar to Singleton?

score 4 · Answer 1

To receive two or more arguments in function, you need to change your signature first.

function pesquisaPagamentos($pesquisa){

for

function pesquisaPagamentos($datainicio, $datafim){

The call should be made this way

pesquisaPagamentos('2015-09-01', '2015-10-05');
pesquisaPagamentos($data1, $data2);

It's no use passing a string separated by a comma, the function will understand that this is just an argument.

Invalid call

pesquisaPagamentos('2015-09-01,2015-10-05');

It is also possible to have two parameters and pass only one value to the function, as long as the second parameter has a default value.

function pesquisaPagamentos($datainicio, $datafim='2015-12-31'){

Call:

pesquisaPagamentos('2015-09-01');

Since php5.6 there is a ... operator that defines multiple parameters for a function, it basically does func_get_args

Example - ideone