将PHP数组传递给JavaScript函数我正在尝试将PHP数组变量转换为JavaScript变量。这是我的密码：<html>
<head>
<script type="text/javascript">
function drawChart(row,day,week,month,date)
{
// Some code...
}
</script>
</head>
<body>
<?php for($counter = 0; $counter<count($au); $counter++)
{
switch($au[$counter]->id)
{
case pageID.'/insights/page_active_users/day':
$day[] = $au[$counter]->value;
break;
case pageID.'/insights/page_active_users/week':
$week[] = $au[$counter]->value;
break;
case pageID.'/insights/page_active_users/month':
$month[] = $au[$counter]->value;
break;
}
}
?>
<script>
drawChart(600/50, '<?php echo $day; ?>', '<?php echo $week; ?>', '<?php echo $month; ?>',
'<?php echo createDatesArray(cal_days_in_month(CAL_GREGORIAN, date('m',strtotime('-1 day')),
date('Y',strtotime('-1 day')))); ?>');
</script>
</body></html>我无法获得PHP数组的值。我该如何解决这个问题？
3 回答
慕仙森
TA贡献1827条经验 获得超8个赞
$php_variable
<script type="text/javascript"> var obj = <?php echo json_encode($php_variable); ?>;</script>
drawChart(600/50, <?php echo json_encode($day); ?>, ...)
JSON.parse(..)
var s = "<JSON-String>";var obj = JSON.parse(s);
侃侃无极
TA贡献2051条经验 获得超10个赞
json_encode
<?php
$phpArray = array(
0 => "Mon",
1 => "Tue",
2 => "Wed",
3 => "Thu",
4 => "Fri",
5 => "Sat",
6 => "Sun",
)?><script type="text/javascript">
var jArray = <?php echo json_encode($phpArray); ?>;
for(var i=0; i<jArray.length; i++){
alert(jArray[i]);
}
</script>
互换的青春
TA贡献1797条经验 获得超6个赞
drawChart(600/50, JSON.parse('<?php echo json_encode($day); ?>'), JSON.parse('<?php echo json_encode($week); ?>'),
JSON.parse('<?php echo json_encode($month); ?>'),
JSON.parse('<?php echo json_encode(createDatesArray(cal_days_in_month(CAL_GREGORIAN, date('m',strtotime('-1 day')),
date('Y',strtotime('-1 day'))))); ?>'))$employee = array( "employee_id" => 10011, "Name" => "Nathan", "Skills" => array( "analyzing", "documentation" => array( "desktop", "mobile" ) ));
{
"employee_id": 10011,
"Name": "Nathan",
"Skills": {
"0": "analyzing",
"documentation": [
"desktop",
"mobile"
]
}}$.ajax({
type: 'POST',
headers: {
"cache-control": "no-cache"
},
url: "employee.php",
async: false,
cache: false,
data: {
employee_id: 10011
},
success: function (jsonString) {
var employeeData = JSON.parse(jsonString); // employeeData variable contains employee array.
});添加回答
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