数据库值为html选择下拉列表中的“已选择”项(Database value as “selected” item in html select dropdown list)

 我认为迄今为止的答案是不正确的。 起作用的是像a *= b这样的复合表达式的评估。 总之， 左边的值是在右边之前计算的 。 来自JLS （强调我的）： 
 
 
 
 在运行时，表达式以两种方式之一进行评估。 
 
 
 如果左边的操作数表达式不是数组访问表达式，则： 
 
 
 首先，评估左手操作数以产生变量。 如果此评估突然完成，则由于相同的原因，分配表达式会突然完成; 右侧的操作数不会被评估，也不会进行分配。 
 
 
 否则， 保存左侧操作数的值，然后评估右侧操作数 。 如果此评估突然完成，则由于同样的原因，分配表达式会突然完成，并且不会进行分配。 
 
 
 否则，左侧变量的保存值和右侧操作数的值将用于执行复合赋值运算符指示的二进制运算。 如果此操作突然完成，则由于同样的原因，赋值表达式会突然完成并且不会发生分配。 
 
 
 否则，二进制运算的结果将转换为左侧变量的类型，并经过值集转换（第5.1.13节）到适当的标准值集（不是扩展指数值集），结果的转换被存储到变量中。 
 
 
 在你的例子中： 
 
a[0] *= a[0] = b[n    ];
 
 
 
 计算a[0]值并存储在tmp 
 
 
 评估a[0] = b[n] ，给出b[n]的值（并且将a[0]的值改为b[n] ） 
 
 
 tmp * a[0] 
 
 
 最后一步的结果被分配给a[0] 
 
 
 所以，你得到的是有效a[0] *= b[n] 。 
 
 编辑：关于从左到右评估分配的困惑：我没有发现在JLS中使用的术语，恕我直言，这是不正确的（尽管它在Java教程中使用）。 它被称为正确的 - **联想*屁股JLS这样说关于任务： 
 
 
 
 有12个分配操作员; 所有都是从句法上右联合的（他们从右到左分组）。 因此，a = b = c意味着a =（b = c），它将c的值赋给b，然后将b的值赋给a。 
 

I think the answers so far are not correct. What comes into play is the evaluation of compound expressions like a *= b. In short, the value of the left hand side is computed before the right hand side. From the JLS (emphasis mine):
 
 
 
At run time, the expression is evaluated in one of two ways.
 
 
If the left-hand operand expression is not an array access expression, then:
 
 
First, the left-hand operand is evaluated to produce a variable. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the right-hand operand is not evaluated and no assignment occurs.
 
 
Otherwise, the value of the left-hand operand is saved and then the right-hand operand is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason and no assignment occurs.
 
 
Otherwise, the saved value of the left-hand variable and the value of the right-hand operand are used to perform the binary operation indicated by the compound assignment operator. If this operation completes abruptly, then the assignment expression completes abruptly for the same reason and no assignment occurs.
 
 
Otherwise, the result of the binary operation is converted to the type of the left-hand variable, subjected to value set conversion (§5.1.13) to the appropriate standard value set (not an extended-exponent value set), and the result of the conversion is stored into the variable.
 
 
In your example:
 
a[0] *= a[0] = b[n    ];
 
 
 
value of a[0] is computed and stored in say tmp
 
 
a[0] = b[n] is evaluated, giving the value of b[n] (and changing the value of a[0] to b[n])
 
 
tmp * a[0] is computed
 
 
the result of the last step is assigned to a[0]
 
 
So, what you get is effectively a[0] *= b[n].
 
EDIT: About the confusion about right-to-left-evaluation of assignments: I did not find that terminology used in the JLS, and IMHO it is not correct (although it is used in the Java tutorials). It is called right-**associative* assThe JLS says this about assignments:
 
 
 
There are 12 assignment operators; all are syntactically right-associative (they group right-to-left). Thus, a=b=c means a=(b=c), which assigns the value of c to b and then assigns the value of b to a.