$result=$menu->show_twoMenu(4);
$col_jobs = array();
$i=0;
while(@$row=mysqli_fetch_array($result)){
$col_jobs[$i] = $row;
$i++;
}
$col_jobs = json_encode($col_jobs);
echo ‘{“coljobs”:’.$col_jobs.’}’;
输出却是：
{
“coljobs”: [
{
*本5文来源gaodai$ma#com搞$$代**码)网@搞代码gaodaima “0”: “5”,
“1”: “清蒸桂鱼”,
“2”: “4”,
“3”: “abc.jpeg”,
“4”: “124”,
“5”: “100”,
“6”: “例”,
“7”: “80”,
“8”: “999”,
“id”: “5”,
“Name”: “清蒸桂鱼”,
“ParentID”: “4”,
“Image”: “abc.jpeg”,
“Discount”: “124”,
“Price”: “100”,
“Unit”: “例”,
“Sale”: “80”,
“Stock”: “999”
}
]
}
为什么会有
“0”: “5”,
“1”: “清蒸桂鱼”,
“2”: “4”,
“3”: “abc.jpeg”,
“4”: “124”,
“5”: “100”,
“6”: “例”,
“7”: “80”,
“8”: “999”,
我不想要，怎么办?
回复讨论(解决方案)
mysqli_fetch_array 返回的是下标数组和关联数组的混合体
按你的需要应使用 mysqli_fetch_assoc 只返回关联数组
转载请注明原文链接：，JSON显示多余的东西
