PHP Notice: Undefined index: step on line 24 Call Stack

Question

When i typed it, this error came out. I want to know why this error happens and how can i fix it. Thank you in advanced.

<html>
<meta charset=utf-8" />
<?php
$dbhandle = @mysql_connect('localhost','root','') or die("Unable to connect to MySQL");
if($sql_banco= mysql_select_db('centroexames',$dbhandle)) {
    echo "Ligação bem sucedida". "<br />". "<br />";
}   else    {
    echo "Ligação mal sucedida". "<br />". "<br />";
};
?>
<body>
<form id="form1" name="form1" method="post" action=""> 
Nomes: 
<select name="candidato"> 
<option value="">Selecione um Nome:</option> 
<?php 
$sql = "select * from candidato";
$resultado = mysql_query($sql);
if($resultado)//teste se houve resultado entra no while 
{ 
while($linhas = mysql_fetch_array($resultado)){ //monta um vetor colocando todos os resultados em $linhas 
?> 
<option value="<?php $linhas['id_candidato'];//aqui é o valor geralmente se coloca o id da tabela ?>" 
<?php if($linhas['id_candidato'] == $_POST['candidato']){ echo "selected"; /*aqui eu testo e vejo se alguma opção foi selecionada eu a mantenho selecionada*/ }?>
<?php echo $linhas['ca_nome'];} } /*aqui é a parte de exibição a informação que o usuario ira ver na tela "as opções"*/ ?> 
</option> 
</select> 
</form> 
</body>
</html>

You didn't post your error, please edit the question so we can help. — Zalomon, Apr 13 '16 at 08:43
Please [edit] your question to include the error and code **as text**. A screenshot of code is not helpful for anyone. — Kirk Beard, Apr 13 '16 at 08:55
@PedroLopes "but the error don't have step" - it does. `Undefined index: >>> step <<< on line 24` On line 24 you're trying to use `$array['step']` and step does not exist in `$array`. The code you posted does not have this, so are you sure you've posted the correct file? Look at line 24 for anything refering to `'step'`. — h2ooooooo, Apr 13 '16 at 09:21

Nana Partykar · Answer 1 · 2016-04-13T09:16:31.173

0

You Missed > closing angle bracket. See my answer.

<option value="" ></option>
                 ^ You missed this

Updated Answer

<form id="form1" name="form1" method="post" action=""> 
    Nomes: 
    <select name="candidato"> 
        <option value="">Selecione um Nome:</option> 
        <?php 
        $sql = "select * from candidato";
        $resultado = mysql_query($sql);
        if($resultado)
        { 
            while($linhas = mysql_fetch_array($resultado)){
            ?> 
            <option value="<?php $linhas['id_candidato']; ?>" 
            <?php if($linhas['id_candidato'] == $_POST['candidato']){ echo "selected"; }?> >
                                                                                           ^ Missed > closing angle bracket 
            <?php echo $linhas['ca_nome'];
            } 
        } ?> 
        </option> 
    </select> 
</form>

edited Apr 13 '16 at 09:16

answered Apr 13 '16 at 09:13

Nana Partykar

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i try that and continuo with the error – Pedro Lopes Apr 13 '16 at 09:15
What error ? @PedroLopes – Nana Partykar Apr 13 '16 at 09:17
the list show me the names of the database but with a error behind. Notice: Undefined index: candidato in C:\wamp64\www\Trabalho\teste.php on line 24 Call Stack #TimeMemoryFunctionLocation 10.0005239632{main}()...\teste.php:0 Paulo Marques – Pedro Lopes Apr 13 '16 at 09:20
@PedroLopes Look at your error. `candidato` does not exist. You're using it here: `$_POST['candidato']`. You're most likely not actually posting this value (hence, the index does not exist). Verify what data you're sending with `print_r($_POST)` and you'll realize your mistake. If it isn't nessesarily available, use `isset()` or `!empty()`. – h2ooooooo Apr 13 '16 at 09:22
thank you @h2ooooooo that help me a lot ;D – Pedro Lopes Apr 13 '16 at 09:45
i put and worked – Pedro Lopes Apr 13 '16 at 09:45
Ok. Good @PedroLopes. *Keep Coding..!!* – Nana Partykar Apr 13 '16 at 10:00
@PedroLopes That won't work. `isset` will return true or false. You need to use `if (!empty($_POST['candidato']) && $linhas['id_candidato'] == $_POST['candidato'])`. – h2ooooooo Apr 13 '16 at 10:07
but worked now only shows the names in the database of the table candidato – Pedro Lopes Apr 13 '16 at 10:20

PHP Notice: Undefined index: step on line 24 Call Stack

1 Answers1