创建 php fom 从表 bd 中选择多个取值 [英] Create php fom select multiple taking values from a table bd

问题描述

表 deportes 有 2 列 4 行:

The table deportes has 2 columns with 4 rows:

<?php
$sql = 'SELECT * FROM deportes';
$resul = mysqli_query($conexion, $sql);

$deportes = array();    
while ($fila = mysqli_fetch_array($resul))
{
     $deportes[] = $fila;
}       
?>

带有选择多个选项的表单:

The form with the select multiple options:

<select name="fan[]" multiple="multiple"> 
    <?php
        foreach ($deportes as $aficion)
        {
            echo "<option value='".$aficion['idD']."'";
            echo " >{$aficion['nombreDep']}  </option>\n";
        }
    ?>            
</select>

从表单中获取值

<?php
if (isset($_POST['fan']))
    {
    $sport = $_POST['fan'];
    }
?>

现在这个

<?php $sport = mysqli_real_escape_string($conexion, $sport); ?>

这种方式将值插入另一个表中

This way insert the values in another table

$idPersona = mysqli_insert_id($conexion);           
$sql = "INSERT INTO mec(id,idD) VALUES ('$idPersona','$sport') ";

结果是我从表 mec 的字段 idD 中得到值0"

And the result is i get the value "0" in the field idD from table mec

推荐答案

如果你print_r你的$_POST['fan']，你会看到这是<强>数组.要获取数组的每个值，您应该使用 for 或 foreach 对其进行迭代:

If you print_r your $_POST['fan'], you will see that this is array. To get every value of array you should iterate over it, with for or foreach:

$idPersona = mysqli_insert_id($conexion);           
foreach ($_POST['fan'] as $sport) {
    echo $sport;   // you will see that now it is string
    $sql = "INSERT INTO mec(id,idD) VALUES ('$idPersona','$sport') ";
    // execute your query
}

当然，您必须转向准备好的语句，以保护您的代码免受 sql 注入.这个问题会给你一个开始.

And of course you must move to prepared statements to protect your code from sql-injection. This question will give you a start.

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