php - 在 while 循环上执行选择查询后，并非所有行都插入到表中 - PHP/MYSQL

Question

我是 stackoverflow 的新手。

我正在执行一个选择查询来填充我需要的输出。当它从查询中获取行时，已获取的每一行都是我将它们插入到特定表中。但是，我需要达到的确切行数是 1,767 行，但在执行查询后，输出为 1759。我缺少 8 行。

我的代码有什么问题？

这是我的代码:

$query2 = "SELECT trihadjustmentitems.AdjID AS `adjid1`, trihadjustment.Adj_ID AS `adjid2`,
trihadjustment.AdjToUnitID AS `adjtounitid`, trihadjustment.AdjDate AS `adjdate`, trihadjustmentitems.InvItemsID AS `invitemid`, 
trihadjustmentitems.SlsItemsID AS `slsitemid`, trihadjustmentitems.RecipeID AS `recipeid`, trihadjustmentitems.Remark AS `remark`,
trihadjustmentitems.AdjQty AS `adjqty`,
trihadjustment.StockCenter_ID AS `stockcenterid1`, trihadjustmentitems.StockCenter_ID AS `stockcenterid2`
FROM trihadjustmentitems
INNER JOIN trihadjustment ON trihadjustmentitems.AdjID = trihadjustment.Adj_ID";

        $result2 = mysqli_query($connection, $query2);


    while($row2 = mysqli_fetch_array($result2))
    {
        $query3 = "INSERT INTO adjustments (adjid1, adjid2, adjtounitid, adjdate, invitemid, slsitemid, recipeid, remark, adjqty, stockcenterid1, stockcenterid2) VALUES ('$row2[adjid1]', '$row2[adjid2]', '$row2[adjtounitid]', '$row2[adjdate]', '$row2[invitemid]', '$row2[slsitemid]', '$row2[recipeid]', '$row2[remark]', '$row2[adjqty]', '$row2[stockcenterid1]', '$row2[stockcenterid2]')";
        $result3 = mysqli_query($connection, $query3);
    }

Answer 1

没有足够的信息来确定任何地方出了问题。您的代码虽然已过时，但不包含任何可以帮助任何人确定问题所在(如果有任何问题)的内容。我们只能在涉及到数字时信任您，而这不是 IT 的运作方式。

根本不需要您的查询。你可以在没有 while 循环的情况下做你想做的事。只需使用 INSERT INTO ... SELECT 语法。

我会用你的代码来说明

$query = <<<EOF
INSERT INTO adjustments 

(adjid1, adjid2, adjtounitid, adjdate, invitemid, slsitemid, recipeid, remark, adjqty, stockcenterid1, stockcenterid2) 

SELECT 
    trihadjustmentitems.AdjID AS `adjid1`, 
    trihadjustment.Adj_ID AS `adjid2`,
    trihadjustment.AdjToUnitID AS `adjtounitid`, 
    trihadjustment.AdjDate AS `adjdate`, 
    trihadjustmentitems.InvItemsID AS `invitemid`, 
    trihadjustmentitems.SlsItemsID AS `slsitemid`, 
    trihadjustmentitems.RecipeID AS `recipeid`, 
    trihadjustmentitems.Remark AS `remark`,
    trihadjustmentitems.AdjQty AS `adjqty`,
    trihadjustment.StockCenter_ID AS `stockcenterid1`, 
    trihadjustmentitems.StockCenter_ID AS `stockcenterid2`

FROM trihadjustmentitems

INNER JOIN trihadjustment ON trihadjustmentitems.AdjID = trihadjustment.Adj_ID

EOF;

$result = mysqli_query($query);

if(!result) {
    echo 'An error occurred';
}