Dynamic table with PHP

Question

Dynamic table with PHP

Asked 9 years, 11 months ago

Viewed 1,270 times

-3

I am creating a dynamic table using PHP.

The problem and the following created a field to modify the values of the variable $conteudo table body however can’t change more than one table field someone there can help me to solve this problem?

PS. to pass the field to be modified and as follows (field 1,field 2,field 3)

<?php

/*
 * CABEÇALHO DA TABELA
 */
$campos = explode(',','id,nome,idade,sexo');
$conteudo = explode(',','1,funano,21,1');


/*
 * MODIFICADOR DE CAMPOS
 */
$modificador = explode(',','21');



echo '<table border="1px" width="100%">';
echo '<tr>';

foreach($campos as $x){
            echo '<th>'.$x.'</th>';
}
echo '</tr>';



/**
 * CONTEUDO DA TABELA
 */
echo '<tr>';
foreach($conteudo as $b){
    /*
     * COMPARANDO O MODIFICADOR COM O CAMPO DA TABELA
     */
    foreach($modificador as $m) {


        if ($b == $m) {
            echo '<th>' . 'campo' . '</th>';
        } else {
            echo '<td>'.$b.'</td>';
        }
    }
}
echo '</tr>';
echo '</table>';

The problem happens when I try to add another field to modify as for example:

$modificador = explode(',','21,funano');

explains more what you tried ... from what I see in the comments who did knows well what has there

– Otto

2014/10/27 at 12:24
runs the script you will understand better

– Alexandre Sousa

2014/10/27 at 12:26
explains better what you need and what you tried to make clear

– Otto

2014/10/27 at 12:30
You can only change one field because it only uses one field in the modifier, it works right here: http://phpfiddle.org/main/code/n7ri-zas8

– Jorge B.

2014/10/27 at 12:31
Jorge B tries to add another field $modifier = explodes(',','21,so-and-so'); and see what will happen to the table

– Alexandre Sousa

2014/10/27 at 12:34
Then you compare "21" from $m with "21" of $b and when this happens it replaces "21" with "field".

– Jorge B.

2014/10/27 at 12:34
perfect now to scold and try to compare more than one table field when I try that the table gets all bugged

– Alexandre Sousa

2014/10/27 at 12:39
This is because you always print either field or value whenever you compare.

– Jorge B.

2014/10/27 at 12:48
And how to solve this jam?

– Alexandre Sousa

2014/10/27 at 12:53
Let’s go continue this discussão in chat.

– Alexandre Sousa

2014/10/27 at 13:25

Show 5 more comments

1 answer

Browser other questions tagged php

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

by Jorge B. • **11,427** points · Answer 1 · 2014-10-27T13:34:12+00:00

The problem is you’re printing the field value or the modifier every time you check all the $m and ends up printing more than once per field.

You can check when there’s match of $m with $b if it’s already been printed:

foreach($conteudo as $b){
    /*
     * COMPARANDO O MODIFICADOR COM O CAMPO DA TABELA
     */
    $isEqual = false;
    foreach($modificador as $m) {

        if ($b == $m) {
            echo '<th>' . 'campo' . '</th>';
            $isEqual = true; // se forem iguais imprime campo
        }
    }
    if(!$isEqual) echo '<td>'.$b.'</td>';//senão imprime o valor de $b
}

phpfiddle

Or you can use the function in_array PHP comparing whether an element is in array:

foreach($conteudo as $b){
    /*
     * COMPARANDO O MODIFICADOR COM O CAMPO DA TABELA
     */
    if (in_array($b, $modificador))
        echo '<th>' . 'campo' . '</th>';
    else
        echo '<td>'.$b.'</td>';
}