Question

我想在我的RPI上建立一个“库存管理器”网站。但是当我输入条形码后，我总是得到productnew.php，当我在数据库中有产品时。（数据库连接并选择在head.php中）

input.php

<head>
    <?php include 'head.php';?>
</head>
<body>
    <header>
    <?php include 'menu.php';?>
</header>
<section>
<?php 
    echo 'Give me a barcode';
    echo '<form method="POST" action="productsearch.php">'; 
    echo '<table>';
    echo '<tr><td>Barcode:</td> <td><input type="text" name="barcode"/></td></tr>';
    echo '<tr><td></td> <td><input type="submit" value="Search"/></td></tr>';
    echo '</table>';
    echo '</form>';
    mysqli_free_result($res);?>             
    </section>
    </body>
    <?php include 'footer.php';?>

productsearch.php

<html><head>
    <?php include 'head.php';?>
</head>
<body>
    <?php 
    $barcode = $_POST["barcode"];
    $sql = "SELECT * FROM inventory WHERE barcode = ' . $barcode . '";
    mysqli_query($conn, $sql) or die ('SQL Error! ' . $sql. ' '. mysqli_error($conn));
    if (mysqli_num_rows($conn, $sql) != 0){
        header("Location: productupdate.php");
        } else {
        header("Location: productnew.php");
    }
include 'footer.php';
?>
</body>
</html>

Answer 1

这个canot是我认为最好的方式，如果你找不到任何最好的解决方案，你可以遵循这个。

<html>
<head>

    <?php include 'head.php';?>

</head>
<body>
    <?php 
    $barcode = $_POST["barcode"];
    $sql = "SELECT * FROM inventory WHERE barcode = ' . $barcode . '";
    mysqli_query($conn, $sql) or die ('SQL Error! ' . $sql. ' '. mysqli_error($conn));
    if (mysqli_num_rows($conn, $sql) != 0){
        header("Location: productupdate.php");
             exit;
        } else {
             header("Location: productnew.php");
             exit;

        }
        include 'footer.php';
     ?>

</body>
</html>

为什么IF ... ELSE ...不工作？

1 个答案: