在使用sequelize-typescript npm
当尝试调用时Street.create(obj)出现错误:
“plainObject”类型的参数不可分配给“Optional<Street, NullishPropertiesOf>”类型的参数。类型“plainObject”缺少类型“Omit<Street, NullishPropertiesOf>”中的以下属性:sequelize、destroy、restore、update 以及其他 39 个属性
这是模型:
import { Table, Model, Column, PrimaryKey } from 'sequelize-typescript';
@Table({ timestamps: false, tableName: 'street' })
class Street extends Model<Street> {
@PrimaryKey
@Column
street_id: string;
@Column
location_id: string;
@Column
location_symbol: string;
@Column
street_name: string;
@Column
street_synonym: string;
@Column
street_symbol: string;
@Column
updated: Date;
}
export default Street;
Run Code Online (Sandbox Code Playgroud)
这是电话
Street.create({
location_id: 'string',
location_symbol: 'string',
street_name: 'string',
street_synonym: 'string',
street_id: 'string',
street_symbol: 'string',
updated: new Date(),
});
Run Code Online (Sandbox Code Playgroud) 我是PHP的新手,遇到了一个我无法解决的问题.
说我们有这个:
<select name="car">
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="mercedes">Mercedes</option>
<option value="audi">Audi</option>
</select>
Run Code Online (Sandbox Code Playgroud)
我的PHP是:
if (isset($_POST['car']) && $_POST['car'] == "Audi") {
echo 'Please select a better car.';
}
Run Code Online (Sandbox Code Playgroud)
这很有效.但是,我在一个数组中生成了一个select:
<select name="q1" value="<?php echo $q1 ?>">
<?php foreach ($toppings as $key => $value) { ?>
<option <?php if ($q1 == $key) { ?>selected="true" <?php }; ?>value="<?php echo $key ?>"><?php echo $value ?></option>
<?php } ?>
</select>
Run Code Online (Sandbox Code Playgroud)
然后我写了这个PHP代码如下:
<?php
$toppings = array(1 => "Anchovie", 2 => "Tomato", 3 => "Corn", 4 => …Run Code Online (Sandbox Code Playgroud)