我當前的代碼,這不符合我的需求,看起來像以下:如何呼應提交輸入表單與特定ID的名稱
<?php
$con = mysqli_connect("localhost", "root", "", "blogs");
$sql = "SELECT * FROM foods order by date desc";
$result = mysqli_query($con, $sql);
while ($r = mysqli_fetch_array($result)) {
//my variables
$buttonname = $r['id']; //for name I want to set on submit input on every echo of form of each results. Ps. My id on my database are set to auto increment.
$lhtitle = $r['title'];
$lhcategory = "foods";
$lhimage = $r['image'];
$lhtext = $r['text'];
echo "<div class='content-content'>";
echo strtoupper("<div class='L-TITLE'>$lhtitle</div>");
echo "<form action='' method='POST'><input type ='submit' value='Add to Collection' name = '$buttonname'";
echo "class='addto-button'></form>";
echo "<div class='content-img'>";
echo "<img style = 'width:100%;' src='images/$lhimage' alt='no image inserted'>";
echo "</div>";
echo "<div class='content-description'><p>$lhtext</p></div>";
echo "</div>";
}
if (isset($_POST[$buttonname])) {
$usn = $_SESSION['logged'];
$sql = "INSERT INTO `$displayuserid` (title,category,image,text) SELECT '$lhtitle','$lhcategory','$lhimage','$lhtext' FROM foods WHERE id = '$buttonname'";
mysqli_query($con , $sql);
}
?>
當我運行這段代碼,只有第一個ID是1
$ buttonname讀取。但我需要所有ID。
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