Question

我知道在一个页面上有两个关于两个查询命令的问题，但我已经尝试了所有内容，但我无法让它工作。

我正在尝试创建一个简单的表单，将输入的值添加到数据库中，然后使用select语句输出数据库。

我做错了什么？

这是我得到的。非常感谢！

<?php
$dbhost = "localhost";
$dbuser = "widget_cms";
$dbpass = "blah123";
$dbdata = "widget_corp";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbdata);

if(mysqli_connect_errno()) {
    die("database connection failed: " .
        mysqli_connect_error() .
            " (" . mysqli_connect_errno() . ")"
    );
} else {
    echo "success!";
} 

?>


<?php 

if (isset($_POST["submit"])) {
    $menu_name = ($_POST["menuName"]);
    $position = ($_POST["pos"]);
    $visibility = ($_POST["vis"]);

    $query1 = "INSERT INTO subjects (menu_name, position, visibility) VALUES ('{$menu_name}', '{$position}', '{$visibility}'); ";

    $result = mysqli_query($connection, $query1);

    if(!$result) {
        die("Database query failed");
    }

} else {
    echo "something didnt work idiot";
}
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
   "http://www.w3.org/TR/html4/loose.dtd">

<html lang="en">
 <head>
  <title>phpMysql</title>
 </head>
 <body>

    <ul>
     <form action="phpMysql.php" method="post">
     Menu Name: <input type="text" name="menuName" value=""><br>
     Position: <input type="text" name="pos" value=""><br>
     Visibility: <input type="text" name="vis" value=""><br>
     <input type="submit" name="submit" value="Create account"><br>
     </form>
    </ul> 

    <?php print_r($_POST); ?>

    <?php

    $query1 = "SELECT * FROM subjects;" ;

    $result = mysqli_query($connection, $query1);

    if(!$result) {
        die("Database query failed");
    }   
    ?>

    <?php while ($row = mysqli_fetch_assoc($result)) {
        var_dump($row);
        echo "<hr />" ;
    }
    ?>

    <?php mysqli_free_result($result); ?>

 </body>
</html>

<?php mysqli_close($connection); ?>

Answer 1

首先，您的编码有几处错误。这一行：

$query1 = "INSERT INTO subjects (menu_name, position, visibility) VALUES ('{$menu_name}', '{$position}', '{$visibility}'); ";

错了。你不需要围绕变量的任何{}。另外，你有一个额外的;在你的报价内。它不会与那些人一起工作。这是正确的代码：

$query1 = "INSERT INTO subjects (menu_name, position, visibility) VALUES ('$menu_name', '$position', '$visibility')";

此行也很糟糕，你还包括一个额外的;在您的查询中。

$query1 = "SELECT * FROM subjects;" ;

应该是这样的：

$query1 = "SELECT * FROM subjects";

你还可以做一些清理工作。很多php块可以合并为单个块，使您的代码看起来更整洁。这是您编辑的代码，包含我注意到的所有修复。

<?php

$dbhost = "localhost";
$dbuser = "widget_cms";
$dbpass = "blah123";
$dbdata = "widget_corp";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbdata);

if(mysqli_connect_errno()) {
    die("database connection failed: " .
        mysqli_connect_error() .
            " (" . mysqli_connect_errno() . ")"
    );
} else {
    echo "success!";
}

if (isset($_POST["submit"])) {
    $menu_name = ($_POST["menuName"]);
    $position = ($_POST["pos"]);
    $visibility = ($_POST["vis"]);

    $query1 = "INSERT INTO subjects (menu_name, position, visibility) VALUES ('$menu_name', '$position', '$visibility')";

    $result = mysqli_query($connection, $query1);

    if(!$result) {
        die("Database query failed");
    }

} else {
    echo "something didnt work idiot";
}
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
   "http://www.w3.org/TR/html4/loose.dtd">

<html lang="en">
 <head>
  <title>phpMysql</title>
 </head>
 <body>

    <ul>
     <form action="phpMysql.php" method="post">
     Menu Name: <input type="text" name="menuName" value=""><br>
     Position: <input type="text" name="pos" value=""><br>
     Visibility: <input type="text" name="vis" value=""><br>
     <input type="submit" name="submit" value="Create account"><br>
     </form>
    </ul> 

<?php

    print_r($_POST);

    $query1 = "SELECT * FROM subjects";

    $result = mysqli_query($connection, $query1);

    if(!$result) {
        die("Database query failed");
    } 

    while ($row = mysqli_fetch_assoc($result)) {
        var_dump($row);
        echo "<hr />" ;
    }

    mysqli_free_result($result);

?>

 </body>
</html>

<?php mysqli_close($connection); ?>

如果您不确定如何正确编码或遇到您不了解的问题，请参阅有关查询的文档。 http://php.net/manual/en/mysqli.query.php我还建议调查准备好的陈述。它们比标准的mysqli查询更好，更安全。 http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

两个数据库查询

1 个答案: