我正在尝试根据值选择数据，我有一个从团队表中选择数据的表单，我需要从选择下拉列表中获取该行的 ID(我可以这样做)，然后使用以确定在执行 AJAX 请求时需要选择的数据。这里是:
团队页面 - 此页面绘制我的图表，该图表通过 AJAX 提取数据来绘制图表及其值:
<form action="teams.php?dashboard_id=<?php echo $dashboard_id; ?>" method="POST">
<select name="teamId">
<option selected="true" disabled="disabled">Please choose...</option>
<?php
$sql = "SELECT * FROM teams WHERE dashboard_id = $dashboard_id";
$result = $conn->query($sql);
if($result->num_rows > 0){
while($row = $result->fetch_assoc()){
echo '<option value=' . $row["team_id"] . '>' . $row["team_name"] . '</option>';
}
}
?>
</select>
<button class="compare">View</button>
</form>
<?php
if (!empty($_POST["teamId"])) {
$teamSelect = $_POST["teamId"];
echo $teamSelect;
}else{
echo "";
}
?>
<div style="max-width: 450px;">
<canvas id="mycanvas" class="container"></canvas>
</div>
这是我在其上执行 SELECT 的 PHP 页面:
<?php
include 'config.php';
$teamID = $_POST['teamId'];
$query = sprintf("SELECT
tm.member_id,
tm.team_id,
m.member_id,
m.firstName,
m.lastName,
m.score_1,
m.score_2,
m.score_3,
m.score_4,
m.score_5,
m.score_6,
m.score_7,
m.score_8,
m.dashboard_id,
t.team_id,
t.team_name,
t.dashboard_id
FROM team_members tm
JOIN members AS m
on m.member_id = tm.member_id
JOIN teams AS t
on t.team_id = tm.team_id
WHERE tm.team_id = '$teamID'"); // This need to be dynamic and got from the POST request on the form above.
$result = $conn->query($query);
$data = array();
foreach ($result as $row) {
$data[] = $row;
}
$result->close();
$conn->close();
header('Content-type: application/json');
print json_encode($data);
?>
我通过另一个文件绘制图表:
$(document).ready(function(){
$.ajax({
url : "http://localhost/acredashAdv/teamData.php",
type : "GET",
success :function(data){
console.log(data);
var chartata = {
labels: [
"Strategic Development and Ownership",
"Driving change through others",
"Exec Disposition",
"Commercial Acumen",
"Develops High Performance Teams",
"Innovation and risk taking",
"Global Leadership",
"Industry Leader"
]};
var ctx = $("#mycanvas");
var config = {
type: 'radar',
data: chartata,
animationEasing: 'linear',
options: {
legend: {
display: true,
position: 'bottom'
},
tooltips: {
enabled: true
},
scale: {
ticks: {
fontSize: 15,
beginAtZero: true,
stepSize: 1
}
}
},
},
LineGraph = new Chart(ctx, config);
var colorArray = [
["#7149a5", false],
["#58b7e0", false],
["#36bfbf", false],
["#69bd45", false],
["#5481B1", false],
["#6168AC", false]
];
for (var i in data) {
tmpscore=[];
tmpscore.push(data[i].score_1);
tmpscore.push(data[i].score_2);
tmpscore.push(data[i].score_3);
tmpscore.push(data[i].score_4);
tmpscore.push(data[i].score_5);
tmpscore.push(data[i].score_6);
tmpscore.push(data[i].score_7);
tmpscore.push(data[i].score_8);
var color, done = false;
while (!done) {
var test = colorArray[parseInt(Math.random() * 6)];
if (!test[1]) {
color = test[0];
colorArray[colorArray.indexOf(test)][1] = true;
done = !done;
}
}
newDataset = {
label: data[i].firstName+' '+data[i].lastName,
borderColor: color,
backgroundColor: "rgba(0,0,0,0)",
data: tmpscore,
};
config.data.datasets.push(newDataset);
}
LineGraph.update();
},
});
});
这一切都很棒，但在我的选择查询中，我需要 WHERE 子句来显示根据团队 ID 的值确定的数据，而不是静态值，但我不确定如何将 ID 从选择传递回AJAX 文件？
最佳答案
然后使用 session 。 $_SESSION['teamID'] = $teamID。然后你就可以在任何地方引用它。请务必使用 session 在每个 php 文件的顶部启动 session 。
session_start();
$teamID = $_POST['teamId'];
$_SESSION['teamID'] = $teamID;
引用文献:
关于php - 根据ID值选择数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41746216/