从mysql读取数据时,php代码没有json输出
我正在尝试使用php以从mysql读取数据时,php代码没有json输出,php,mysql,json,Php,Mysql,Json,我正在尝试使用php以json格式检索mysql数据,下面是我的代码,但执行后我得到的是空白输出,如果我使用var_dump()检查数组元素的输出,那么我可以看到记录,但作为一个整体,json输出没有显示,由于我不熟悉json,因此无法排除故障,如果有人能检查并找到问题,这将是一款很棒的thnx <?php header('Content-Type: application/json'); $dbhost = 'localhost'; $dbuser = 'root'; $dbpass =
json
格式检索mysql数据
,下面是我的代码,但执行后我得到的是空白输出,如果我使用var_dump()
检查数组元素的输出,那么我可以看到记录,但作为一个整体,json
输出没有显示,由于我不熟悉json
,因此无法排除故障,如果有人能检查并找到问题,这将是一款很棒的thnx
<?php
header('Content-Type: application/json');
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'mypassword';
$dbname = 'mydbname';
//setting records limit per page is 15
$rec_limit = 15;
//Establishing Connection
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* Get total number of records */
$sql = "SELECT count(*) FROM ImageUploads";
$retval = $conn->query($sql);
if(! $retval )
{
die($mysqli->error.__LINE__);
}
$rec_count = $retval->fetch_row();
$rec_count = $rec_count[0];
// Checking for page parameter to set.
if( isset($_GET{'page'} ) )
{
$page = $_GET{'page'} + 1;
$offset = $rec_limit * $page ;
}
else
{
$page = 0;
$offset = 0;
}
//getting all data from table
$sql = "SELECT * FROM ImageUploads ORDER BY slno DESC ".
"LIMIT $offset, $rec_limit";
$retval = $conn->query($sql);
if(! $retval )
{
die($mysqli->error.__LINE__);
}
//creating an array for response
$response = array();
if ($retval->num_rows > 0) {
$response["wallpapers"] = array();
$response["success"] = 1;
$response["count"]= $rec_count;
while ($row = $retval->fetch_array()) {
// temp wallpaper array
$wallpaper = array();
$wallpaper["id"] = $row["slno"];
$wallpaper["orig_url"] = "http://doupnow.com/AndroidApp/ImageUploads/".$row["image_url"];
$wallpaper["thumb_url"]="No Thumb";
$wallpaper["downloads"] = $row["downloads"];
$wallpaper["fav"] = $row["views"];
// push all data into final response array
array_push($response["wallpapers"], $wallpaper);
}
// echoing JSON response
echo str_replace('\/','/',json_encode($response,JSON_PRETTY_PRINT));
} else {
// no wallpapers found
$response["success"] = 0;
$response["message"] = "No Wallpapers found";
}
mysqli_close($conn);
?>
看起来你的大脑有问题 $wallpaper[“原始url”]=”.$row[“图像url”];
从中删除$row[“image\u url”]并对其进行测试。问题已解决,只是一个简单的问题,感谢上帝。我把这句话改成--
现在我得到了json输出。这可能会显示在
$response
和var\u dump
also@YashParekh............the附加了var_dump($response)的输出。2天后,您可以接受自己的答案,这将有助于将来阅读此问题的其他人
array(3) {
["wallpapers"]=>
array(1) {
[0]=>
array(5) {
["id"]=>
string(1) "1"
["orig_url"]=>
string(95) "http://doupnow.com/AndroidApp/ImageUploads/raman.das@gmail.com_sOw$vN8T_IMG_20171203_200638.jpg"
["thumb_url"]=>
string(8) "No Thumb"
["downloads"]=>
string(1) "0"
["fav"]=>
string(1) "0"
}
}
["success"]=>
int(1)
["count"]=>
string(1) "1"
}
echo str_replace('\/','/',json_encode($response));
//echo str_replace('\/','/',json_encode($response,JSON_PRETTY_PRINT));