HTML PHP concatenation

Question

HTML PHP concatenation

Asked 5 years ago

Viewed 54 times

0

Below follows the code and has a comment where I’m having trouble concatenating html + php... basic silly error but already merged the cuckoo and do not see the problem. thank you.

<?php
              // Monta consulta MySQL para o gráfico
              $sqlano = "SELECT DISTINCT YEAR(dateent) as Year FROM `cadastro`";
              $queryano = $mysqli->query($sqlano);

?>

            <form method="GET" action="" name="menuForm">     
              <header>
                  <h3>Relatório de pedidos por mês do ano de: </h3>
                        <select id="ano" name="ano" onchange="document.forms['menuForm'].submit();">
                            <option value="2000" disabled selected> Escolha o ano </option> 

                                <?php
                                   while($dadosano = $queryano->fetch_array()){
                                   //essa linha abaixo estou com problema na concatenação mas já fundi a cabeça e não encontro o erro. 
                                   echo "<option value='".$dadosano['Year']."'  ". isset($_GET['ano']) && $_GET['ano']== $dadosano['Year']?' selected="selected"' : ' ' ;  "  > ".$dadosano['Year']. "</option>";

                                    }

                                 ?>

                          </select> 


              </header>
            </form>

1

And what would that be ; in the middle of the line, just after the ternary condition?

– Woss

2019/09/25 at 16:07
is to end the condition... condition ? code : code;

– Deadoc

2019/09/25 at 16:12
1

Not only are you ending the condition, you’re ending your echo; everything after will give syntax error (probably the error you got and omitted in the question).

– Woss

2019/09/25 at 16:14
No error in execution but did not do what you have to do, popular select and take what was selected...

– Deadoc

2019/09/25 at 17:07

1 answer

Browser other questions tagged php html

You are not signed in. Login or sign up in order to post.

by Woss • **73,416** points · Answer 1 · 2019-09-25T16:18:54+00:00

The error is syntax. Do:

echo "<option value='".$dadosano['Year']."'  ". isset($_GET['ano']) && $_GET['ano']== $dadosano['Year']?' selected="selected"' : ' ' ;  "  > ".$dadosano['Year']. "</option>";

How did you use a ; in the middle of the expression, it would be the same as:

echo "<option value='".$dadosano['Year']."'  ". isset($_GET['ano']) && $_GET['ano']== $dadosano['Year']?' selected="selected"' : ' ';
"  > ".$dadosano['Year']. "</option>";

This second line is not a valid PHP.

Other than that, there’s a lot of responsibility for just one expression. Simplify this and you won’t have so many problems.

$selected = (isset($_GET['ano']) && $_GET['ano'] == $dadosano['Year']) ? 'selected="selected"' : '';
echo "<option value='{$dadosano['Year']}' {$selected}>{$dadosano['Year']}</option>";