我有这段代码:
public function updateSegmentGender ($product_id, $segment_gender) {
$this->connect();
$product_id = $this->escapeString($product_id);
$row_count = count($segment_gender);
for ($row = 0; $row < $row_count; $row++) {
$this->select('product_seg_gender', '*', NULL,
'product_id = "'.$product_id.'" AND gender = "'.$segment_gender[$row][0].'"', NULL, NULL);
$res = $this->getResult();
$res = count($res);
$gender = $segment_gender[$row][0];
$status = $segment_gender[$row][1];
if ($res <> 0) {
// UPDATE
$data = array ('status'=>$status);
$this->update('product_seg_gender', $data, 'product_id = "'.$product_id.'" AND gender = "'.$gender.'"');
}else {
// INSERT
$data = array ('product_id'=>$product_id, 'gender'=>$gender, 'status'=>$status);
$this->insert('product_seg_gender', $data);
}
}
}
我正在使用这样的方法:
$user = new Product($db_server, $db_user, $db_password, $db_name);
$user->connect();
$segment_gender = array ( array ("all", "active"),
array ("female", "active"));
$res = $user->updateSegmentGender ('303', $segment_gender);
print_r($res);
但为什么我总是收到此错误消息:
Warning: Cannot use a scalar value as an array in /home/***/public_html/class/Database.class.php on line 130
但是,数据库已成功更新。我做错了什么?
更新:这是Database.class.php的完整第97-145行
// Function to SELECT from the database
public function select($table, $rows = '*', $join = null, $where = null, $order = null, $limit = null){
// Create query from the variables passed to the function
$q = 'SELECT '.$rows.' FROM '.$table;
if($join != null){
$q .= ' JOIN '.$join;
}
if($where != null){
$q .= ' WHERE '.$where;
}
if($order != null){
$q .= ' ORDER BY '.$order;
}
if($limit != null){
$q .= ' LIMIT '.$limit;
}
// echo $table;
$this->myQuery = $q; // Pass back the SQL
// Check to see if the table exists
if($this->tableExists($table)){
// The table exists, run the query
$query = $this->myconn->query($q);
if($query){
// If the query returns >= 1 assign the number of rows to numResults
$this->numResults = $query->num_rows;
// Loop through the query results by the number of rows returned
for($i = 0; $i < $this->numResults; $i++){
$r = $query->fetch_array();
$key = array_keys($r);
for($x = 0; $x < count($key); $x++){
// Sanitizes keys so only alphavalues are allowed
if(!is_int($key[$x])){
if($query->num_rows >= 1){
$this->result[$i][$key[$x]] = $r[$key[$x]];
}else{
$this->result[$i][$key[$x]] = null;
}
}
}
}
return true; // Query was successful
}else{
array_push($this->result,$this->myconn->error);
return false; // No rows where returned
}
}else{
return false; // Table does not exist
}
}
答案 0 :(得分:1)
注意:我希望您需要$res = $user->updateSegmentGender ('303', $segment_gender);
需要$data
值,并根据假设我使用$data
作为返回部分。
可能是由于$data
未初始化。我在尝试将变量$data
声明为数组之前,在使用它之前我还要进行另一项更改,function updateSegmentGender()
需要返回部分,所以我把它放了。
public function updateSegmentGender ($product_id, $segment_gender) {
$this->connect();
$product_id = $this->escapeString($product_id);
$row_count = count($segment_gender);
$data = array();//Initialize variable...
for ($row = 0; $row < $row_count; $row++) {
$this->select('product_seg_gender', '*', NULL,
'product_id = "'.$product_id.'" AND gender = "'.$segment_gender[$row][0].'"', NULL, NULL);
$res = $this->getResult();
$res = count($res);
$gender = $segment_gender[$row][0];
$status = $segment_gender[$row][1];
if ($res <> 0) {
// UPDATE
$data = array ('status'=>$status);
$this->update('product_seg_gender', $data, 'product_id = "'.$product_id.'" AND gender = "'.$gender.'"');
}else {
// INSERT
$data = array ('product_id'=>$product_id, 'gender'=>$gender, 'status'=>$status);
$this->insert('product_seg_gender', $data);
}
}
//Return funal value in array format...
return $data;
}