我想显示数据库中的图像。我的图像字段的数据类型是BLOB。我在本网站的其他一些问题中使用了代码,但代码不起作用。我想在Google地图的信息窗口中显示图片。根据屏幕截图,图像标签正在显示,但图像本身不是。
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我的代码:
<?php
require "connect.php";
$sql = "SELECT * FROM tblLocation";
$result = mysqli_query($conn,$sql);
$count = mysqli_num_rows($result);
echo "<script>
function initMap() {
var infowindow = new google.maps.InfoWindow();
var uluru = {lat: 14.617766, lng: 121.057154};
var map = new google.maps.Map(document.getElementById('map'), {
zoom: 4,
center: uluru
});";
if ($count > 0){
while ($row = mysqli_fetch_array($result)) {
echo "var content = '<div class=".'content'.">'+
'<img src=".'data:image/jpeg;base64,'.base64_encode($row['image'])."'/>'+
'<p>".$row['address']."</p>'+
'<span class=".'label'.">Branch:</span>'+
'<span class=".'details'.">".$row['branch']."</span>'+
'<span class=".'label'.">Province:</span>'+
'<span class=".'details'.">".$row['province']."</span>'+
'<span class=".'label'.">Phone:</span>'+
'<span class=".'details'.">".$row['phone']."</span>'+
'<span class=".'label'.">Latitude:</span>'+
'<span class=".'details'.">".$row['lat']."</span>'+
'<span class=".'label'.">Longitude:</span>'+
'<span class=".'details'.">".$row['lng']."</span>'+
'<button class=".'btn-edit'.">Edit</button>'+
'<button class=".'btn-delete'.">Delete</button>'+
'</div>';";
echo "var marker = new google.maps.Marker({
content:content,
position:{ lat:{$row['lat']}, lng:{$row['lng']} },
map: map
});
marker.addListener('click', function(event) {
infowindow.open( map, this );
infowindow.setContent( this.content );
infowindow.setPosition( event.latLng );
}.bind( marker ));";
}
}
echo "}
</script>";
?>