php - mysql 数据库的更新逻辑问题

Question

我有以下 PHP 表单，它回发到 mysql 数据库。我的问题是更新查询似乎有效，但总是被“checked”覆盖。我想要做的是检查从数据库中获取当前值，然后如果 post 中有值，则获取该值。现在...为什么这不起作用？检查它是否在 _POST 中时是否需要有 else 子句？如果是这种情况，我是否需要使用 $checkDeleted = ""; 初始化变量？

<?php
error_reporting(E_ALL);
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"];
}
$checkDeleted = "";
$con = mysqli_connect("localhost","user","pw", "db");
$getformdata = $con->query("select ARTICLE_NO, deleted from STATUS where ARTICLE_NO = '$pk'");
while ($row = mysqli_fetch_assoc($getformdata)) {
    $ARTICLE_NO = $row['ARTICLE_NO'];
    $checkDeleted = $row['deleted'];
}
$checkboxes = (isset($_POST['checkboxes'])? $_POST['checkboxes'] : array());
if (in_array('deleted', $checkboxes)) $checkDeleted = 'checked';
if($cmd=="submitinfo") {
    if ($ARTICLE_NO == null) {
        $statusQuery = "INSERT INTO STATUS VALUES (?, ?)";
        if ($statusInfo = $con->prepare($statusQuery)) {
            $statusInfo->bind_param("ss", $pk, $checkDeleted);
            $statusInfo->execute();
            $statusInfo->close();
        } else {
            print_r($con->error);
        }
    } else if ($ARTICLE_NO == $pk) {
        $statusQuery = "UPDATE STATUS SET deleted = ? WHERE ARTICLE_NO = ?";
        if ($statusInfo = $con->prepare($statusQuery)) {
            $statusInfo->bind_param("ss", $checkDeleted, $pk);
            $statusInfo->execute();
            $statusInfo->close();
        } else {
            print_r($con->error);
        }
    }
}
if($cmd=="EditStatusData") {
    echo "<form name=\"statusForm\" action=\"test.php?pk=".$pk."\" method=\"post\" enctype=\"multipart/form-data\">
                        <input type=\"checkbox\" name=\"checkboxes[]\" value=\"deleted\" ".$checkDeleted." />
                        <label for=\"deleted\">Delete</label>
                        <input type=\"hidden\" name=\"cmd\" value=\"submitinfo\" />
                        <input name=\"Submit\" type=\"submit\" value=\"submit\" />
        </form>";
}
?>

我尝试更改该行以将 checkDeleted 设置为以下内容，这没有什么区别......尽管应该如此？

if (in_array('deleted', $checkboxes)) {
$checkDeleted = 'checked';
} else {
$checkDeleted = '';
}

编辑:好的，我已经设法让它工作，但只有在更改为

之后

$checkDeleted = in_array('已删除', $checkboxes) ? “已检查”:“”；

按照下面的答案，但这仍然不起作用。为了让它工作，我必须删除数据库查询，并将其替换为submitinfo分支中的一个和EditStatusData分支中的一个......为什么？为什么不能只有一个查询？

if($cmd=="submitinfo") {

$getformdata = $con->query("select ARTICLE_NO from STATUS where ARTICLE_NO = '$pk'");

while ($row = mysqli_fetch_assoc($getformdata)) {
    $ARTICLE_NO = $row['ARTICLE_NO'];
}    
    if ($ARTICLE_NO == null) { etc

和

if($cmd=="EditStatusData") {
$getformdata = $con->query("select deleted from STATUS where ARTICLE_NO = '$pk'");

while ($row = mysqli_fetch_assoc($getformdata)) {
    $checkDeleted = $row['deleted'];
} etc

Answer 1

这与您的其他问题几乎相同

mysql not updating from php form

代码没有任何问题，它完全按照你想要的方式工作

What I want to do is get the current value from the database, and then if there is a value in post, get that instead.

情况1:不带勾的html表单

1. 从数据库读取 $checkDeleted = 'checked'
2. 如果 $_POST['checkboxes']['deleted'] 未设置，则保留 $checkDeleted 不变
3. 将“已检查”写入数据库

案例2.带勾的html表单

1. 从数据库读取 $checkDeleted = 'checked'
2. 如果设置了 $_POST['checkboxes']['deleted']，则更改 $checkDeleted = 'checked'
3. 将“已检查”写入数据库

所以无论你是否打勾，一旦你将数据库值改为checked，就无法更改它了

我会假设您想要做的始终是使用复选框设置的内容覆盖数据库值，在这种情况下

替换此行

 if (in_array('deleted', $checkboxes)) $checkDeleted = 'checked';

有了这个

 $checkDeleted = in_array('deleted', $checkboxes) ? 'checked' : '';