Php 如何从json输出中获取确切的对象
我有一个json编码的输出,比如Php 如何从json输出中获取确切的对象,php,json,Php,Json,我有一个json编码的输出,比如 "[{"COLUMN_NAME":"htag"},{"COLUMN_NAME":"title"},{"COLUMN_NAME":"sounds"},{"COLUMN_NAME":"recx"}]" 我想在数组中单独获取列名,并在同一个php文件中回显它 代码: 这就是我所做的 $sql=mysql_query("SELECT `COLUMN_NAME` FROM `INFORMATION_SCHEMA`.`COLUMNS` WHERE `TABLE_SCH
"[{"COLUMN_NAME":"htag"},{"COLUMN_NAME":"title"},{"COLUMN_NAME":"sounds"},{"COLUMN_NAME":"recx"}]"
我想在数组中单独获取列名,并在同一个php文件中回显它
代码:
这就是我所做的
$sql=mysql_query("SELECT `COLUMN_NAME`
FROM `INFORMATION_SCHEMA`.`COLUMNS`
WHERE `TABLE_SCHEMA`='dbname'
AND `TABLE_NAME`='tbname'");
while($row=mysql_fetch_assoc($sql))
$out[] = $row;
$reds = json_encode($out);
我想在数组中单独获取列名
零钱:
$sql=mysql_query("SELECT `COLUMN_NAME`
FROM `INFORMATION_SCHEMA`.`COLUMNS`
WHERE `TABLE_SCHEMA`='dbname'
AND `TABLE_NAME`='tbname'");
while($row=mysql_fetch_assoc($sql))
$out[] = $row['COLUMN_NAME'];
//Column names are now in the array $out.
$reds = json_encode($out);
顺便问一下,你不应该使用mysql。为什么会这样depreciations@user2992174看这里,告诉我你是模糊的医生