我有一个PHP表单,我用来输入数据到数据库(phpmyadmin),我用SELECT查询来显示数据库中的所有值,以在php窗体中查看。从PHP创建的JSON提供错误的数据?
另外我还有另一个PHP文件,我用它从相同的数据库表创建JSON。
这里的时候,我进入国外的语言,如“Experienţapersonală:”使用保存在DB值是“ExperienÈ>一personală:‘但是当我使用select查询,显示这同一的PHP表单,它的到来正确’Experienţapersonală: ”。因此,DB是正确的,现在使用下面的PHP代码来创建JSON
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "aaps";
// Create connection
$con=mysqli_connect($servername,$username,$password,$dbname);
// Check connection
mysqli_set_charset($con, 'utf8');
//echo "connected";
$rslt=mysqli_query($con,"SELECT * FROM offers");
while($row=mysqli_fetch_assoc($rslt))
{
$taxi[] = array('code'=> $row["code"], 'name'=> $row["name"],'contact'=> $row["contact"], 'url'=> $row["url"], 'details'=> $row["details"]);
}
header("Content-type: application/json; charset=utf-8");
echo json_encode($taxi);
?>
和JSON看起来像
[{"code":"CT1","name":"Experien\u00c8\u203aa personal\u00c4\u0192: ","contact":"4535623643","url":"images\/offers\/event-logo-8.jpg","details":"Experien\u00c8\u203aa personal\u00c4\u0192: jerhbehwgrh 234234 hjfhjerg#$%$#%#4"},{"code":"ewrw","name":"Experien\u00c8\u203aa personal\u00c4\u0192: ","contact":"ewfew","url":"","details":"eExperien\u00c8\u203aa personal\u00c4\u0192: Experien\u00c8\u203aa personal\u00c4\u0192: Experien\u00c8\u203aa personal\u00c4\u0192: "},{"code":"Experien\u00c8\u203aa personal\u00c4\u0192: ","name":"Experien\u00c8\u203aa personal\u00c4\u0192: ","contact":"","url":"","details":"Experien\u00c8\u203aa personal\u00c4\u0192: "}]
在这个“\ u00c8 \ u203aa”这是错了,它应该是“\ u021b“(t)。
因此,用于创建JSON使这个问题。
但我无法准确找到它为什么会这样。请帮助
确定数据库中的数据正确吗?在编码之前尝试对'$ taxi'数组执行'var_dump'。 – Jerodev
是的数据库是正确becoz当我试图做相同的PHP文件选择查询是越来越正确的date.ok我会做,让你知道 – Bangalore
array(4){[0] => array(5){[“code” ] =>字符串(3)“CT1”[“name”] =>字符串(29)“ExperienâðpersonalÔÆ':“[”contact“] =>字符串(10)”4535623643“[”url“] => string(30)“images/offers/event-logo-8.jpg”[“details”] => string(66)“ExperienúºpersonalÔÆ':jerhbehwgrh 234234 hjfhjerg#$%$#%#4 “} [1] =>数组(5){[”code“] =>字符串(4)”ewrw“ – Bangalore