Question

设置变量并在start.php中正确显示所有内容。我可以在表单中的两个显示变量之间进行选择。

问题是：选中并且提交表单后，变量$projektcheck或$projektcheckk不会传递给nxtfile.php。当我使用print($_POST['prjkttype']);检查时，变量未定义且没有任何内容被打印。它会立即切换到我的echo"Data is wrong";

为什么它没有被传递到另一个文件，虽然它被设置和选择？

start.php：

<?PHP 
    session_start();
    $projektcheck = $_SESSION['projektname'];
    $projektcheckk = $_SESSION['projektnamee'];                
?>
 [other codestuff]
<script>
function validateForm() {
    var fields = ['prjkttype']

    var i, l = fields.length;
    var fieldname;
    for (i = 0; i < l; i++) {
        fieldname = fields[i];
    }   
    f1.method="POST";
    f1.submit();
}
</script>


[other codestuff]

<form action="nxtfile.php" method="POST" name="f1">
    <select class="selectstyle" name="prjkttype" form="submit">
<?PHP 
    if (isset($projektcheck)) {
        echo"<option value='1'>$projektcheck </option>";    
    } else {
    }  

    if (isset($projektcheckk)) {
        echo"<option value='2'>$projektcheckk </option>";   
    } else {
    }  
?>
    </select>

    <input type=button name=s1 value="Insert" onClick="validateForm()">

</form>
[other codestuff]

nxtfile.php：

<?PHP
session_start();
include ("../connection.php");
$prjkttype=$_POST['prjkttype'];


print($_POST['prjkttype']); 

$checkindb="SELECT * FROM projekte WHERE (projektname = '$prjkttype' )";
$dbcheckk=mysqli_query($con ,$checkindb);
$dbcheckedd=mysqli_fetch_array($dbcheckk);

if($dbcheckedd > 0){
    header ("Location: xy.php");
} else {
    echo "Data is wrong.";
}
?>

Answer 1

您需要将值设置为您的变量，如下所示。它将传递值而不是变量。这将有效：

<template name="OneInvoice">
    <p>Invoice id: {{invoice._id}}</p>
    <p>Invoice total: {{invoice.total}}</p>
    {{#let company=(loadCompany invoice.companyId)}}
        <p>Company name: {{company.name}}</p>
        <p>Company business address: {{company.businessAddress}}</p>
    {{/let}}
</template>

通过选择表单将变量从文件传递到文件

1 个答案: