While循环-其他输入中只有一个输入正在通过POST发送值 [英] While loop - Only one input from many others is sending a value through POST

问题描述

这是我从称为optionale的表中获取输入名称和值并对其进行以下处理的代码：

 < form role = form autocomplete = off action = includes / functions / fisa-init.php method = POST> 
<？php 
 connectDB（）; 
 $ query = mysqli_query（$ mysqli， SELECT * FROM`optionale`）或die（mysqli_error（$ mysqli））; 
 while（$ row = mysqli_fetch_array（$ query））
 {
？> 
< span><？php echo $ row [’denumire’]; < / span> 
< input type = text name = nrBucati> 
< input type = hidden value =<？php echo $ row [’cod’];？> name = codProdus> 
<？php}？> 
< / form> 
  

可选表如下：

HTML看起来像这样：

如上图所示，我使用HTML，输入的名称（取自optionale表）和我在其中写入值的实际输入。

fisa-init.php：

  $ stmt3 = $ mysqli-> prepare（ 
更新
`stocuri` 
 SET 
`cantitate` =`cantitate`-？
 WHERE`cod` =？
）； 
 
 
 $ stmt3-> bind_param（ is，$ bucata，$ cod）; 
 
 //设置参数并执行
 $ bucata = mysqli_real_escape_string（$ mysqli，$ _POST [’nrBucati']）; 
 $ cod = mysqli_real_escape_string（$ mysqli，$ _POST [’codProdus’]）; 
 
 if（！$ stmt3-> execute（））
 {
 echo Exercuția aîntâmpinato eroare：（。$ stmt3-> errno。） 。 $ stmt3->错误; 
} 
 $ stmt3-> close（）; 
 
 $ mysqli-> close（）; 
  

在上面的代码（fisa-init.php）中，我试图从中获取所有输入值我的HTML并更新另一个名为 stocuri 的表中的行：

祝你有美好的一天

This is the code where I get my input names and values from a table called optionale and doing something with these:

<form role="form" autocomplete="off" action="includes/functions/fisa-init.php" method="POST">
<?php
   connectDB();
   $query = mysqli_query($mysqli, "SELECT * FROM `optionale`") or die(mysqli_error($mysqli));
   while($row = mysqli_fetch_array($query))
   { 
?>
   <span><?php echo $row['denumire']; ?></span>
   <input type="text" name="nrBucati">
   <input type="hidden" value="<?php echo $row['cod']; ?>" name="codProdus">
<?php } ?>
</form>

The optionale table looks like this:

The HTML looks like this:

As you can see in the last picture, I have in HTML, a name for input (taken from optionale table) and the actual input in which I write a value.

fisa-init.php:

$stmt3 = $mysqli->prepare("
            UPDATE 
            `stocuri` 
            SET 
            `cantitate` = `cantitate` - ?
            WHERE `cod` = ?
            ");


    $stmt3->bind_param("is", $bucata, $cod);

    // set parameters and execute
    $bucata = mysqli_real_escape_string($mysqli, $_POST['nrBucati']);
    $cod = mysqli_real_escape_string($mysqli, $_POST['codProdus']);

    if (!$stmt3->execute()) 
        {
            echo "Execuția a întâmpinat o eroare: (" . $stmt3->errno . ") " . $stmt3->error;
        }
    $stmt3->close();

    $mysqli->close();

In the code above (fisa-init.php) I am trying to take all the input values from my HTML and update rows in another table called stocuri:

As you can see, only the second row from stocuri table was updated, but I wrote values in all 5 inputs. It got only the last input.

How to modify the while loop in order to take all my inputs value?

If something is not clear, I apologize a hundred times. I will explain all the informations that are needed.

P.S. cod in table optionale is the same with cod in table stocuri. In this way, I know where to update values.

解决方案

Each <input> MUST have an individual name or a named array.

So give each an aditional number like name1,name2 or use an named array like name[]

Finally this name="codProdus[]" is your solution.

Read more here HTML input arrays

Have a nice day

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