为什么我的代码中出现以下错误?
致命错误:无法在写入上下文中使用函数返回值。
刚开始使用此函数时,我在if ... else中有return语句,该错误似乎与return语句有关。但是我用echo语句替换了它们,但仍然出现错误,所以我不知道发生了什么。任何帮助或建议?
public function wallPostComments() {
// This function processes wall post comments
// pull the submitted comment data
$returnedPostId = $this->inuput->post('entryId');
$returnedCommentData = $this->input->post('returnedCommentData');
// pull the required session data
$userid = $this->session->userdata('userid');
// select the sql data from wallPosts and wallPostComments
$query = $this->db->query("SELECT * FROM wallPosts, wallPostComments");
// loop through the mysql rows and process the expanded sql code
foreach ($query->result() as row()) {
if($row->idwallPosts == $JSONedIdWallPosts) {
echo "success";
} else {
echo "failure";
}
}
}
最佳答案
row()
什么时候成为函数?
尝试:
foreach ($query->result() as $row) {
关于php - 致命错误:无法在以下情况的写入上下文中使用函数返回值:,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/9985698/