总体思路是绘制一个div中某人得分的图表。我有一个按钮，当点击它时运行图形绘制功能。我还有另一个函数，它使用switch语句从我的数据库中检索数据，因为该函数与其他按钮共享。
我的数据检索功能：
var getdata = function(button_id) {
$.ajax({
type: "POST",
url: "../scripts/getdata.php",
dataType: "html",
data: {id: button_id},
success: function(result){
$("#profilebox").html(result);
}
});
};
运行getdata.php并将值返回到空白div。
getdata.php：
<?php
session_start();
$switchcase = $_POST['id'];
$email = $_SESSION['user']['email'];
//connect to database here
$result=mysqli_query($con,"SELECT * FROM users WHERE email = '$email'");
switch ($switchcase) {
case "profile_home":
while($row = mysqli_fetch_array($result)) {
echo $row['username'] . "'s Profile<br><br>";
echo "Name: " . $row['firstname'] . ' ' . $row['lastname'] . "<br><br>";
echo "Things I like:<br>";
echo $row['like'] . "<br><br>";
echo "Things I dislike:<br>";
echo $row['dislike'] . "<br><br>";
echo "Other Sports:<br>";
echo $row['sports'];
};
break;
case "profile_scores":
while($row = mysqli_fetch_array($result)) {
$row['correct'];
$row['incorrect'];
};
break;
case "profile_merits":
//CODE GOES HERE;
break;
case "profile_help":
//CODE GOES HERE;
break;
case "profile_edit":
//CODE GOES HERE;
break;
}
mysqli_close($con);
?>
接收POSTed div id（profile_scores），从数据库获取数据，切换到第二种情况。现在问题出在这里，我不知道如何通过
$row['correct'];
$row['incorrect'];
返回原始页面并让它们显示在图表
中/* correct value */
和
/* incorrect value */
是
图形绘制功能：
function drawVisualization() {
getdata("profile_scores");
// Create and populate the data table.
var data = google.visualization.arrayToDataTable([
['', 'Correct', 'Incorrect'],
['Scores', /* correct value */, /* incorrect value */],
]);
var options = {
'title': 'Total Scores Overall',
'width': 600,
'height': 400,
'hAxis': {title: ''},
'backgroundColor': 'transparent'
};
// Create and draw the visualization.
new google.visualization.ColumnChart(document.getElementById('profilebox')).
draw(data, options);
};
此功能是用户单击按钮时运行的函数，因此它调用getdata函数，然后使用从getdata.php接收的值绘制图形。
感谢任何帮助：）
答案 0 :(得分：0)
您的PHP代码可以在（JSON编码的）关联数组中编码返回参数。然后，您可以通过ajax响应中的键分别检索它们，并在JavaScript代码中传递它们。
答案 1 :(得分：0)
我完全不确定你要求的是什么，但这是我的猜测。
如果您希望从页面上的数据库中 $ 2 中获取 $ result 。然后，您可以使用JSON将PHP数组传递给Javascript，然后完成剩下的工作。就像假设你的switch语句没有2那样
case "profile_scores":
{
$row = $result->fetch_array(MYSQLI_NUM);
echo json_encode($row);
break;
}
在您的ajax响应中执行此操作
$.ajax({
type: "POST",
url: "../scripts/getdata.php",
dataType: "html",
data: {id: button_id},
success: function(result){
console.log(result);
console.log(JSON.parse(result));
var phparray = JSON.parse(result);
var correct = phparray[12]; //<---This will return 100
var incorrect = phparray[13]; //<---This will return 10
//I Dont know which one is your correct or incorrect column number but you can have an idea now
}
});
这样你就得到了你的PHP数组到javascript，这是数据库结果。另外，我建议您更正Switch语法。这不会产生太大的影响，但应该小心，这是正确的语法。
switch($switchcase)
{
case "case1":
{
//Code to go
break;
}
case "case2":
{
//Code to go
break;
}
case "case3":
{
//Code to go
break;
}
}
请注意，我已将 break; 命令放在切换案例
中答案 2 :(得分：0)
尝试使用json.php为javascript调用获取json解析字符串。 下载链接：class.json.php
Javascript的变化：
var getdata = function(button_id) {
$.ajax({
type: "POST",
url: "../scripts/getdata.php",
dataType: "json", // html -> json
data: {id: button_id},
success: function(result){
$("#profilebox").html(result.html);
return result;
}
});
};
function drawVisualization() {
var data = getdata("profile_scores");
// Create and populate the data table.
var data = google.visualization.arrayToDataTable([
['', 'Correct', 'Incorrect'],
['Scores', data.correct, data.incorrect],
]);
var options = {
'title': 'Total Scores Overall',
'width': 600,
'height': 400,
'hAxis': {title: ''},
'backgroundColor': 'transparent'
};
// Create and draw the visualization.
new google.visualization.ColumnChart(document.getElementById('profilebox')).
draw(data, options);
};
getdata.php中的更改：
<?php
// INCLUDE JSON
// ----------------------------
include_once("class.json.php");
$json = new JSON();
$coreJSON = array();
// ----------------------------
session_start();
$switchcase = $_POST['id'];
$email = $_SESSION['user']['email'];
// call as result.html in javascript
$coreJSON['html'] = "";
// call as result.correct in javascript
$coreJSON['correct'] = "";
// call as result.incorrect in javascript
$coreJSON['incorrect'] = "";
//connect to database here
$result=mysqli_query($con,"SELECT * FROM users WHERE email = '$email'");
switch ($switchcase) {
case "profile_home": // <- CHANGES
while($row = mysqli_fetch_array($result)) {
$coreJSON['html'].= <<<EOF
{$row['username']}'s Profile<br /><br />
Name: {$row['firstname']} {$row['lastname']}<br /><br />
Things I like:<br>
{$row['like']} <br /><br />
Things I dislike:<br />
{$row['dislike']}<br /><br />
Other Sports:<br />
{$row['sports']}
EOF;
}
/* Start the closing EOF at the beginning of the line, and delete all spaces and tabs after the semicolon. Next code must start in a new line.*/
break;
case "profile_scores":
while($row = mysqli_fetch_array($result)) {
$coreJSON['correct'].= $row['correct']; // <- CHANGES
$coreJSON['incorrect'].= $row['incorrect']; // <- CHANGES
}
break;
case "profile_merits":
//CODE GOES HERE;
break;
case "profile_help":
//CODE GOES HERE;
break;
case "profile_edit":
//CODE GOES HERE;
break;
}
mysqli_close($con);
// CLOSING AND CONVERTING
// ----------------------------
$encodeJSON = $json->encode($coreJSON);
header('Content-Type: text/json'); // text/plain is good to
header('Content-Length: '.strlen($encodeJSON));
die($encodeJSON);
// ----------------------------
?>
我希望它有所帮助。