这在这个 php 脚本中意味着什么？ $行[$i]=&$行[$i];看来 $row 正在获取自身的引用，这没有意义

Question

表格结构：8列简单员工信息表

CREATE TABLE `emp2` (
    `empno` int(11) NOT NULL,
    `ename` text,
    `job` text,
    `mgr` int(11) DEFAULT NULL,
    `hiredate` text,
    `sal` double DEFAULT NULL,
    `comm` double DEFAULT NULL,
    `deptno` int(11) DEFAULT NULL,
    PRIMARY KEY (`empno`),
    UNIQUE KEY `pk_emp` (`empno`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

测试 csvfile：测试 emp.csv 文件

7369,SMITH,CLERK,7902,"1980-12-23 00:00:00",800,,20
7499,ALLEN,SALSMAN,7698,"1980-12-23 00:00:00",1600,300,30
7521,WARD,SALESMAN,7698,"1980-12-23 00:00:00",1250,500,30

php 脚本：我删除了大部分异常代码以使其更短 我可以理解 array_unshift($row ,$fmt) 是让 $row 如下所示，所以有效数据是从位置 1 到 8 。

array(9) {
   [0] =>
   string(8) "ssssssss"
   [1] =>
   string(4) "7934"
   [2] =>
   string(6) "MILLER"
   [3] =>
   string(5) "CLERK"
   [4] =>
   string(4) "7782"
   [5] =>
   string(19) "1980-12-23 00:00:00"
   [6] =>
   string(4) "1300"
   [7] =>
   string(0) ""
   [8] =>
   string(2) "10"
}

<?php
$tname = 'emp';
$fname = 'emp.csv';
$rownum = 0;
function create_insert_stmt($table, $ncols)
{//create insert sql
    $stmt = "insert into " . $table . " values(";
    foreach (range(1, $ncols) as $i) {
        $stmt .= " ? ,";
    }
    $stmt = preg_replace("/,$/", ');', $stmt);
    return $stmt;
}

try {
    $db = new mysqli('localhost', 'root', 'root', 'scott');
    $db->autocommit(FALSE);
    $res = $db->prepare("select * from " . $tname);
    $ncols = $res->field_count;
    $res->free_result();
    $ins = create_insert_stmt($tname, $ncols);
    $fmt = str_repeat("s", $ncols);
    $res = $db->prepare($ins);
    $fp = new SplFileObject($fname, "r");
    while ($row = $fp->fgetcsv()) {
        if (strlen(implode('', $row)) == 0) continue;
        array_unshift($row, $fmt);//put 'ssssssss' into array $row
        foreach (range(1, $ncols) as $i) {
            $row[$i] =& $row[$i];
            //what does this mean ,when i remove & why it can`t work
        }
        call_user_func_array(array(&$res, "bind_param"), $row);
        //the original code is :
        //call_user_func_array(array(&$res,"bind_param"),&$row);
        //it seems i can`t use &$row in php5.6,so i removed & then it worked fine
        $res->execute();

        $rownum++;
    }
    $db->commit();
    print $rownum . "rows inserted into " . $tname . "\n<br>";

} catch (Exception $e) {
    print "exception:";
    die($e->getMessage() . '\n<br>');
}

Answer 1

根据：http://php.net/manual/en/mysqli-stmt.bind-param.php

面向对象风格

bool mysqli_stmt::bind_param ( 字符串 $types , 混合 &$var1 [, 混合 &$... ] ) 程序风格

bool mysqli_stmt_bind_param（mysqli_stmt $stmt，字符串$types，混合&$var1 [，混合&$...]）

bind_param函数的参数都是引用，

所以，很明显 $row[$i]=&$row[$i] 正在获取自身的引用。

令我困惑的是：获取自身的参考到底是什么？ 是写时复制吗？或者初始化内存中的数组？ $row是由$fp->getcsv()生成的，所以它应该已经存在于内存中