Question

我有以下表格，我尽可能地减少了表格，但没有确定问题的来源。我试图将表单值插入数据库。

然而，当尝试使用下面的表格时，它会吐出：

查询为空

两次。 STATUS表存在，删除和通知的字段也存在。该表目前没有内容。我不明白为什么查询会为空？

<?php
if (isset($_GET["cmd"]))
  $cmd = $_GET["cmd"]; else
if (isset($_POST["cmd"]))
  $cmd = $_POST["cmd"]; else
die("Invalid URL");
if (isset($_GET["pk"])) {
    $pk = $_GET["pk"];
}
if (isset($_POST["deleted"])) {
    $deleted = $_POST["deleted"];
}
if (isset($_POST["notice"])) {
    $notice = $_POST["notice"];
}
$con = mysqli_connect("localhost","user","password", "db");
if (!$con) {
    echo "Can't connect to MySQL Server. Errorcode: %s\n". mysqli_connect_error();
    exit;
}
$con->set_charset("utf8");
$getformdata = $con->query("select * from STATUS where ARTICLE_NO = '$pk'");
$checkDeleted = "";
$checkNotice = "";
while ($row = mysqli_fetch_assoc($getformdata)) {
    $checkDeleted = $row['deleted'];
    $checkNotice = $row['notice'];
}
if($cmd=="submitinfo") {
    $statusQuery = "INSERT INTO STATUS VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
    if ($statusInfo = $con->prepare($userQuery)) {
        $statusInfo->bind_param("ss", $deleted, $notice);
        $statusInfo->execute();
        $statusInfo->close();
        echo "true";
    } else {
        echo "false";
    }
    print_r($con->error);
}
if($cmd=="EditStatusData") {
    echo "<form name=\"statusForm\" action=\"test.php\" method=\"post\" enctype=\"multipart/form-data\">
<h1>Editing information for auction: ".$pk."</h1>
Löschung Ebay:
<input type=\"checkbox\" name=\"deleted\" value=\"checked\" ".$checkDeleted." />
<br />
Abmahnung:
<input type=\"checkbox\" name=\"notice\"  value=\"checked\" ".$checkNotice."  />
<br />
<input type=\"hidden\" name=\"cmd\" value=\"submitinfo\" />
<input name=\"Submit\" type=\"submit\" value=\"submit\" />
</form>";
} else {
    print_r($con->error);
}

Answer 1

这里有拼写错误：

$statusQuery = "INSERT INTO STATUS VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
if ($statusInfo = $con->prepare($userQuery)) {

第二行不是$ statusQuery吗？

Answer 2

尝试将一些内容放入表格中...如果没有任何内容可以返回，则会告诉您查询为空。

Answer 3

您的作业是

$ statusQuery =“INSERT INTO ....

但您引用了一个名为$ userQuery

的变量

if（$ statusInfo = $ con＆gt; prepare（ $ userQuery ））{

混合表单和mysql数据库查询的问题

3 个答案: