PHP 数组格式错误
PHP Array in wrong format
我有一个查询,我希望将结果插入数组中,所以毕竟我会将其编码为 JSON,但我的问题是我希望数据像这样设置:
array[0] = project1, project2, project3;
array[1] = item1, item2, item3;
我有这个:
array[0] = project1;
array[1] = project2;
array[2] = project3;
等等..
这是我目前所做的:
$info = array();
$items = mysql_query("SELECT * FROM `vision`.`projects` WHERE proj_area = 'area_1'");
if (mysql_num_rows($items) != 0) {
while($proj = mysql_fetch_array($items)) {
$proj_name = $proj['proj_name'];
$proj_beg = $proj['proj_beg'];
$proj_end = $proj['proj_end'];
array_push($info, $proj_name, $proj_beg, $proj_end );
}
}
echo json_encode($info);
我的查询结果给了我这些结果:
["nome", "0000-00-00", "0000-00-00", "Projeto 2", "2016-12-12", "2020-07-30", "Projeto", "2017-02-03", "2018-03-10"]
这是我的 $.getJSON 代码:
$.getJSON("includes/get_area.php",function(data){
console.log(data);
})
我做错了什么?
试试这个;这将在三个数组索引中的每一个中添加一个列表。
$info = array();
$items = mysql_query("SELECT * FROM `vision`.`projects` WHERE proj_area = 'area_1'");
if (mysql_num_rows($items) != 0) {
while($proj = mysql_fetch_array($items)) {
$info[0][] = $proj['proj_name'];
$info[1][] = $proj['proj_beg'];
$info[2][] = $proj['proj_end'];
}
}
echo json_encode($info);
我有一个查询,我希望将结果插入数组中,所以毕竟我会将其编码为 JSON,但我的问题是我希望数据像这样设置:
array[0] = project1, project2, project3; array[1] = item1, item2, item3;
我有这个:
array[0] = project1; array[1] = project2; array[2] = project3;
等等..
这是我目前所做的:
$info = array();
$items = mysql_query("SELECT * FROM `vision`.`projects` WHERE proj_area = 'area_1'");
if (mysql_num_rows($items) != 0) {
while($proj = mysql_fetch_array($items)) {
$proj_name = $proj['proj_name'];
$proj_beg = $proj['proj_beg'];
$proj_end = $proj['proj_end'];
array_push($info, $proj_name, $proj_beg, $proj_end );
}
}
echo json_encode($info);
我的查询结果给了我这些结果:
["nome", "0000-00-00", "0000-00-00", "Projeto 2", "2016-12-12", "2020-07-30", "Projeto", "2017-02-03", "2018-03-10"]
这是我的 $.getJSON 代码:
$.getJSON("includes/get_area.php",function(data){
console.log(data);
})
我做错了什么?
试试这个;这将在三个数组索引中的每一个中添加一个列表。
$info = array();
$items = mysql_query("SELECT * FROM `vision`.`projects` WHERE proj_area = 'area_1'");
if (mysql_num_rows($items) != 0) {
while($proj = mysql_fetch_array($items)) {
$info[0][] = $proj['proj_name'];
$info[1][] = $proj['proj_beg'];
$info[2][] = $proj['proj_end'];
}
}
echo json_encode($info);