我有一個非常簡單的PHP表單,它顯示一個複選框,並將存儲如果它被檢查或不在數據庫中。這適用於初始插入,但不適用於更新。我測試了$ saleid等於$ pk的情況,並且它沒有輸入if分支來更新...爲什麼?mysql不從php格式更新
<?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"]; }
$checkfield = "";
$checkboxes = (isset($_POST['checkboxes'])? $_POST['checkboxes'] : array());
if (in_array('field', $checkboxes)) $checkfield = 'checked';
$con = mysqli_connect("localhost","user","", "db");
if (!$con) { echo "Can't connect to MySQL Server. Errorcode: %s\n". mysqli_connect_error(); exit; }
$con->set_charset("utf8");
$getformdata = $con->query("select saleid, field from STATUS where saleid = '$pk'");
$saleid = "";
while ($row = mysqli_fetch_assoc($getformdata)) {
$saleid = $row['saleid'];
$checkfield = $row['field'];
}
if($cmd=="submitinfo") {
if ($saleid == null) {
$statusQuery = "INSERT INTO STATUS VALUES (?, ?)";
if ($statusInfo = $con->prepare($statusQuery)) {
$statusInfo->bind_param("sssssssssssss", $pk, $checkfield);
$statusInfo->execute();
$statusInfo->close();
} else {
print_r($con->error);
}
} else if ($saleid == $pk) {
$blah = "what";
$statusQuery = "UPDATE STATUS SET field = ? WHERE saleid = ?";
if ($statusInfo = $con->prepare($statusQuery)) {
$statusInfo->bind_param("ss", $checkfield, $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\">
<h1>Editing information for Auction No: ".$pk."</h1>
<input type=\"checkbox\" name=\"checkboxes[]\" value=\"field\" ".$checkfield." />
<label for=\"field\">Test</label>
<br />
<input type=\"hidden\" name=\"cmd\" value=\"submitinfo\" />
<input name=\"Submit\" type=\"submit\" value=\"submit\" />
</form>";
}
?>
也記住,如果複選框不打勾,它不會在php – bumperbox 2009-05-25 00:13:16
的複選框[]數組中返回一個值混合工作正常插入...只是不更新... – Josh20002 2009-05-25 01:28:32