Question

我写了这段代码

$serial_no = 1; 
while($rowss = mysqli_fetch_array($selectQuery))
echo $rowss['work_date'];
$mauzaNameQry = mysqli_query($conn,"select moza_name from moza_names where          mauza_id =1");
echo mysql_error();
$mauzaNameRslt = mysqli_fetch_array($mauzaNameQry);
`echo $rowss['work_date'];
`

当我写了另一个select内部循环查询时，它没有返回$ rowss []数组中的任何内容，因为我已经在上面写了选择查询内部while循环，它从moza_name表中选择数据。如果删除moza_name表的内部查询，$ rowss []会显示数据。它甚至没有显示任何错误。

Answer 1

我不确定你要做什么，但这可能是吗？

<?php
    $serial_no = 1;
    while($row = mysqli_fetch_array($selectQuery)) {
        echo $row['work_date'];
    }
    $mauzaNameQry = mysqli_query($conn,"SELECT moza_name FROM moza_names WEHRE `mauza_id`='1'") or die(mysqli_error($mauzaNameQry));
    while($row = mysqli_fetch_array($mauzaNameQry)) {
        echo $row['work_date'];
    }
?>

更新：

您是否尝试进行嵌套循环？

    <?php
    $serial_no = 1; 
    while($rowss = mysqli_fetch_array($selectQuery)) {
    echo $rowss['work_date'];
        $mauzaNameQry = mysqli_query($conn,"SELECT moza_name FROM moza_names where `mauza_id`='1'") or die();
        while($mauzaNameRslt = mysqli_fetch_array($mauzaNameQry)) {
            echo $mauzaNameRslt['work_date'];
        }
    }
?>

更新3：

<?php
        $userId = 1; //ID to search for
        $ID = mysqli_query($conn,"SELECT * FROM work_sheet where `mauza_id`='$userId'") or die(); //get ID and if it doesn't exist, exit
        $IdFromTable1 = $ID['moza_id']; // set the correct ID from table1
        $mauzaNameQry = mysqli_query($conn,"SELECT * FROM moza_names where `mauza_id`='$IdFromTable1'") or die(); //check for table1's correct ID on table2
        echo $mauzaNameRslt['moza_name']; //echo the name from table1's ID
?>

mysql查询在循环内部不起作用

1 个答案: