将按钮的名称和值存储在 mysql 表行中
[英]Store name and value of a Button in a mysql table row
问题描述
I've been hanging on this problem for 3 days and just can't solve it.
我已经在这个问题上挂了 3 天，但无法解决它。 It seems relatively simple, but unfortunately, I can't do it, because I'm a noob.看起来比较简单，可惜我做不到，因为我是菜鸟。
I would like to use the button's onclick function to get its 2 pieces of information我想使用按钮的 onclick 函数来获取它的 2 条信息
```html
<input id="'.$row['id'].'" type="button" value="Pause" name="'.$row['name'].'" onclick="pushDataToDB()">```
- name = "'. $ row [' name '].'" [the name reference from a database table] name = "'.$row [' name'].'" [来自数据库表的名称引用]
- and value="Pause"和价值=“暂停”
this two informations i would like to store in some other database table.这两个信息我想存储在其他一些数据库表中。 I have tried to solve this problem in different ways.我试图以不同的方式解决这个问题。
I have tried to store in different ways the event.target.name & event.target.event javascript output in PHP Variables and then use it in the mysql insert line but i failed.我试图以不同的方式在 PHP 变量中存储 event.target.name 和 event.target.event javascript 输出，然后在 mysql 插入行中使用它，但我失败了。
I have tried to post the value to the ajax.php then store the value in a PHP variable and use this as Value to push it to the database, but this also doesn't work我试图将值发布到 ajax.php 然后将该值存储在 PHP 变量中并将其用作值将其推送到数据库，但这也不起作用
索引.php
<?php
include_once ('dbh.php');
if(isset($_POST['name'])){
$sqlAufPause = "INSERT INTO aktuellaufpause (name, pausenart) VALUES ('$name', 'BP')";
}
if ($conn->query($sqlAufPause) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
function pushDataToDB() { $.get("ajax.php"); return false; } <?php include_once ('dbh.php'); if(isset($_POST['name'])){ $sqlAufPause = "INSERT INTO aktuellaufpause (name, pausenart) VALUES ('$name', 'BP')"; } if ($conn->query($sqlAufPause) === TRUE) { echo "New record created successfully"; } else { echo "Error: " . $sql . "<br>" . $conn->error; } ?> I would say this is the reason, why the name Value is empty, but i don't know how to fix ist..我会说这就是为什么名称 Value 为空的原因，但我不知道如何修复它。
2 个解决方案
解决方案1
1 2020-10-23 07:42:34
You can use JQuery + AJAX to perform asynchronous POST您可以使用 JQuery + AJAX 执行异步 POST
<?php
$con = mysqli_connect("localhost", "root", "", "testdb");
$query = mysqli_query($con, "SELECT * FROM mytable WHERE id = 1");
$row = mysqli_fetch_assoc($query);
print_r($row);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</head>
<body>
<div class="button2">
<form id="frm">
<input id="<?php echo $row['id'] ?>" type="button" value="Pause" name="<?php echo $row['name'] ?>" onclick="pushDataToDB()">
</Form>
</div>
</body>
<script type="text/javascript">
function pushDataToDB() {
var name = "<?php echo $row['name'] ?>";
var value = "Pause";
$.ajax({
url: "ajax.php",
type: "POST",
data: {
"name": name,
"value": value
},
success: function(e) {
if (e == "1") {
alert("Success!");
} else {
alert("error!");
}
}
});
}
</script>
</html>
ajax.php ajax.php
<?php
$con = mysqli_connect("localhost", "root", "", "testdb");
$name = $_POST['name'];
$value = $_POST['value'];
$query = "INSERT INTO mytable(name, value) values('$name','$value')";
$result = mysqli_query($con, $query);
if ($result) {
echo "1";
} else {
echo "Error!";
}
EDITTTT****编辑部****
I changed我变了
var name = "<?php echo $row['id'] ?>";
to到
var name = "<?php echo $row['name'] ?>";
since it is the "name" you want stored in the database因为它是您想要存储在数据库中的“名称”
解决方案2
1 2020-10-24 13:41:25
I solved the problem this way.我是这样解决问题的。 Thank you so much!!!非常感谢！！！
index.php索引.php
<?php
$sql = "SELECT * FROM `DB-TABLENAME` ORDER BY `VALUE1`, `VALUE2`";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_assoc($result)){
echo "<tr>";
echo "<td>";
echo '<input style="margin: 0 auto 6px 17px;" type="checkbox" id="scales" name="scales">';
echo "</td>";
echo "<td>";
echo $row['VALUE1'];
echo "</td>";
echo "<td>";
echo $row['VALUE2'];
echo "</td>";
echo "<td>";
echo '<div class="button1">
<input id="'.$row['id'].'" type="button" value="BP" name="'.$row['VALUE2'].'" onclick="pushDataToDB()">
</div>';
echo "</td>";
echo "<td>";
echo '<div class="button2">
<form id="frm">
<input id="'.$row['id'].'" type="button" value="Mittagspause" name="'.$row['name'].'" onclick="pushDataToDB()">
</form>
</div>';
echo "</td>";
echo "</tr>";
}
} else {
echo "there are no comments";
}?>
<script type="text/javascript">
function pushDataToDB() {
var name = event.target.name;
var value = event.target.value;
$.ajax({
url: "ajax.php",
type: "POST",
data: {
"name": name,
"value": value
},
success: function(e) {
if (e == "1") {
alert("Success!");
} else {
alert("error!");
}
}
});
}
</script>
ajax.php ajax.php
<?php
include_once('dbconnectioncredentials.php');
$con = mysqli_connect($servername, $username, $password, $dbname);
$name = $_POST['VALUE1'];
$value = $_POST['VALUE2'];
$query = "INSERT INTO aktuellaufpause (VALUE1, VALUE2) VALUES ('$name','$value')";
$result = mysqli_query($con, $query);
if ($result) {
echo "1";
} else {
echo "Error!";
}
?>
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