我在使用jquery显示图像时遇到问题。从数据库中检索图像,我想使用jquery显示它。下面是我的代码
<?php
require_once("db.php");
if(isset($_POST['input'])):
$id = $_POST['id'];
$selectImage = mysqli_prepare($link,"SELECT product_image FROM products WHERE id = ? ");
mysqli_stmt_bind_param($selectImage, "i",$id);
mysqli_stmt_execute($selectImage);
mysqli_stmt_bind_result($selectImage,$image);
mysqli_stmt_fetch($selectImage);
if($image):
echo $image;
else:
echo "no image to display";
endif;
endif;
?>
function imgViews(){
$(".imageView").click(function(){
var id = $(this).attr("id");
$.ajax({
url: 'class/display_image.php',
type: 'POST',
data: {input: 'input',
id:id
},
success:function(data){
****display image*****
},
dataType: "text"
});
});
}
imgViews();
<div id="imageContainer>
<img src="img/....">
</div>
答案 0 :(得分:0)
尝试一下
success:function(data){
$('#imageContainer').html('<img src="'+data+'"/>');
},
答案 1 :(得分:0)
index.php
<?php
require_once("db.php");
if(isset($_POST['input'])):
$id = $_POST['id'];
$selectImage = mysqli_prepare($link,"SELECT product_image FROM products WHERE id = ?
");
mysqli_stmt_bind_param($selectImage, "i",$id);
mysqli_stmt_execute($selectImage);
mysqli_stmt_bind_result($selectImage,$image);
mysqli_stmt_fetch($selectImage);
if($image):?>
<div id="img_id"><?php echo $image; ?></div>
<?php
else: ?>
<div id="img_id"></div>
<?php
endif;
endif;
?>
ajax.php
function imgViews(){
$(".imageView").click(function(){
var id = $('img_id').text().trim();
if(id != ""){
$.ajax({
url: 'class/display_image.php',
type: 'POST',
data: {input: 'input',
id:id
},
success:function(data){
$('#imageContainer').html('<img src="'+data+'"/>');
},
dataType: "text"
});
}
});
}
imgViews();
答案 2 :(得分:0)
我希望这会对您有所帮助
$(“#imageContainer img”)。attr(“ src”,data);
答案 3 :(得分:0)
尝试一下
$('#pic').html('"<img src="'+ picURL + data +'"/>');