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[英]Changing 2nd dropdown values according to 1st using javascript values coming from database
[英]JavaScript does not echo values coming from database
我首先從數據庫中獲取值沒有問題。 然后,我可以回顯這些值,但是在這些值變為空之后。 值以某種方式沒有超過這一點。 這是我的代碼。
PHP和MySQL的一部分
$rows = array();
$result = mysql_query("SELECT * FROM kayitlar");
$i=1;
while($row = mysql_fetch_array($result)) {
$rows []= array(
'id' => $row['id'],
'ad' => $row['ad'],
'saat' => $row['saat'],
);
$i++;
}
到此為止沒有問題。 這是我遇到問題的其余代碼
<script type="text/javascript">
window.onload = function () {
var chart = new CanvasJS.Chart("chartContainer",
{
title:{
text: "title"
},
animationEnabled: true,
axisY: {
title: "Zaman (saat)"
},
legend: {
verticalAlign: "bottom",
horizontalAlign: "center"
},
theme: "theme2",
data: [
{
type: "column",
showInLegend: true,
legendMarkerColor: "grey",
legendText: "saat",
dataPoints: [
{y:<?php echo json_encode($row['ad']); ?>, label: "<?php echo json_encode($row['saat']); ?> "},
]
}
]
});
chart.render();
}
</script>
這里我卡住<?php echo json_encode($row['ad']); ?>
<?php echo json_encode($row['ad']); ?>
沒有任何價值
您的$ rows數組索引數組包含每個索引中的鍵。 因此,您需要從數組中提取鍵值對。
while循環完成后,添加以下代碼
$arr['ad'] = array_column($rows,"ad");
$arr['saat'] = array_column($rows,"saat");
$arr['id'] = array_column($rows,"id");
現在在您的jS中使用它
<script type="text/javascript">
window.onload = function () {
var chart = new CanvasJS.Chart("chartContainer",
{
title:{
text: "title"
},
animationEnabled: true,
axisY: {
title: "Zaman (saat)"
},
legend: {
verticalAlign: "bottom",
horizontalAlign: "center"
},
theme: "theme2",
data: [
{
type: "column",
showInLegend: true,
legendMarkerColor: "grey",
legendText: "saat",
dataPoints: [
{y:<?php echo json_encode($arr['ad']); ?>, label: "<?php echo json_encode($arr['saat']); ?> "},
]
}
]
});
chart.render();
}
</script>
您的數組是multidimensional array
因此array_column
使用array_column
從每個索引中獲取specific column
值並應用json_encode()
<?php echo json_encode(array_column($rows,'ad')); ?>
<?php echo json_encode(array_column($rows,'saat')); ?>
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