Mysqli查询不起作用但没有错误

Question

I'm trying to insert data into db from php

public function writeIntoDB($fields, $values, $table){
    $query = $this->composeInsertQuery($fields, $values, $table);
    if (($this->conn->query($query)) === TRUE){
    ...
    }
    echo "Error: " . query . "<br>" . $this->conn->error;
}

I wrote this function but the code inside the if statement is never executed, I get only the echo without error because $this->conn->error return null. The other function called is composeInsertQuery(...):

private function composeInsertQuery($fields, $values, $table){
    $query = "INSERT INTO " . $table . " (id,";
    foreach ($fields as $value) {
        $query .= $value . ",";
    }
    $query = substr($query, 0, -1) . ") VALUES ('$this->key',";
    foreach ($values as $value) {
        $query .= "'$value',";
    }
    $query = substr($query, 0, -1) . ")";
    return $query;
}

It compose the query in a modular way because I need to have different number of fields every time. However this is not the problem because I tried to print the composed query and assign it directly to a variable, like this:

public function writeIntoDB($fields, $values, $table){
    $query = "INSERT INTO ... all the query ....";
    if (($this->conn->query($query)) === TRUE){
    ...
    }
    echo "Error: " . query . "<br>" . $this->conn->error;
}

But even in this case I got the echo with $this->conn->error always empty string. Of course the same query works correctly from terminal.

$this->conn->query return NULL and this is weird according to PHP documentation:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN > queries mysqli_query() will return a mysqli_result object. For other > successful queries mysqli_query() will return TRUE.

Answer 1

删除composeInsertQuery（）中的id字段