现在我正在尝试编辑我的数据库表的字段，但问题是当我单击它时选择第 1 行，当我第二次单击它时它选择第 2 行，依此类推，无论我选择的是哪一行。
为了便于理解；当我点击第 1 行时，它选择了第一行，但是当我再次点击第 1 行时，它选择了第 2 行并显示了要编辑的第 2 行数据，当我点击第 3 行时，它显示了要编辑的第 3 行数据和儿子。它处于一种循环中。
下面是我的代码
<!doctype HTML>
<head>
<link rel="stylesheet" href="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/themes/smoothness/jquery-ui.css">
<link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
<link rel="stylesheet" href="css/venview.css">
</head>
<body>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.js"></script>
<input type="button" id="addbtn" value="Add">
<?php
include 'includes/head.php';
include 'ven_connect.php';
include "dashboard.php";
if (isset($_GET["page"]))
{
$page = (int) $_GET["page"];
}
else
{
$page=1;
};
$start_from = ($page-1) * 4;
$result = mysqli_query($conn , "SELECT * FROM vendor LIMIT $start_from, 4") or die (mysqli_error ($conn));
echo "<table title='Vendors'>";
echo '<tr>';
echo "<th>Sr</th>";
echo "<th>Edit</th>";
echo "<th>Delete</th>";
echo "<th>Name</th>";
echo "<th>PhoneNo</th>";
echo "<th>Email</th>";
echo "</tr>";
echo "<tr>";
while($row = mysqli_fetch_array( $result )) {
// display the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['id'] . '</td>';
echo '<td>' . '<img src="edit.png" style = "height:35px;margin-left :8px;" onclick = "edit()">' . '</td>';
echo '<td>' . '<img src="delete.png" style = "height:35px;margin-left :8px;" onclick = "deleterecord('.$row['id'].')">' . '</td>';
echo '<td>' . $row['Name'] . '</td>';
echo '<td>' . $row['Number'] . '</td>';
echo '<td>' . $row['email'] . '</td>';
echo "</tr>";
echo '<div id="dialog-confirm" style="display:none;">';
echo'<form method="post" action="edit_ven.php" class="ajax">';
echo'<label for="id">ID</label>';
echo'<input type="text" id="id" name="id" value= '.$row['id'].' readonly> ';
echo'<label for="name">Name</label>';
echo'<input type="name" id="name" name="name" value= '.$row['Name'].'>';
echo'<label for="number">Number</label>';
echo'<input type="number" id="number" name="number" value= '.$row['Number'].'>';
echo'<label for="email">Email</label>';
echo'<input type="email" id="email" name = "email" value= '.$row['email'].'>';
echo'</form>';
echo'</div>';
}
echo "</tr>";
echo "</table>";
//Pagination!!
if($page > 1)
{
$prev= $page - 1;
echo " <a href='{$_SERVER['PHP_SELF']}?page=$prev'>Prev</a> ";
}
$result = mysqli_query($conn , "SELECT * FROM vendor") or die (mysqli_error ($conn));
$total_records = mysqli_num_rows($result);
$total_pages = ceil($total_records / 4);
$range = 3;
for ($x = ($page - $range); $x <($page + $range); $x++) {
if (($x > 0) && ($x <= $total_pages)) {
if ($x == $page) {
echo " [<b>$x</b>] ";
}
else {
echo " <a href='{$_SERVER['PHP_SELF']}?page=$x'>$x</a> ";
}
}
}
if($page != $total_pages)
{
$nextpage=$page+1;
echo " <a href='{$_SERVER['PHP_SELF']}?page=$nextpage'>Next</a> ";
}
?>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.10.3/jquery-ui.min.js"></script>
<script src="smoke.js"></script>
<link rel="stylesheet" href="css/venview.css">
<script type="text/javascript">
function deleterecord ( id ) {
smoke.confirm("Do you want to delete?", function(result){
if(result)
{
window.location.href = 'delete_ven.php?id=' + id;
}
else{
header("Location: ven_view.php");
}});
}
function edit() {
$('#dialog-confirm').dialog({
modal: true,
width: 400,
height: 400,
buttons: {
update: function() {
window.location.href = 'edit_ven.php?id=' + id;
},
Cancel: function() {
$( this ).dialog( "close" );
}
}
});
}
</script>
</body>
</html>
最佳答案
您应该发送一个 id 到编辑功能，然后发送 html 代码中的设置。
function edit(id)
关于php - 没有选择正确的行进行编辑，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31229367/