php - mysqli_fetch_array()中的警告

为什么我的数组由于mysqli_fecth数组而出错？我可以对代码进行的处理是正确的，因为我是一个初学者。我将代码更改为mysqli_fetch_assoc（），但错误相同。

这是我做的代码：

      <?php
      $query = mysqli_query($con,"SELECT t.po_nomor, p.nama_supplier, p1.nama_barang , t.total FROM
        (SELECT t1.po_nomor, (SUM(t1.jumlah)-SUM(COALESCE(t4.terima,0))over(PARTITION by t4.refrence) ) as total
        FROM pengiriman_supply t1
        INNER JOIN data_supplier t2 ON t1.idsupplier = t2.id_supplier
        INNER JOIN data_barang t3 ON t1.idbarang = t3.idbarang
        LEFT JOIN masuk t4 ON t4.refrence = t1.po_nomor
        where t1.tanggal BETWEEN date_sub(curdate(), interval 120 day) AND curdate()
        group by t1.po_nomor,t4.po_nomor) t
        INNER JOIN pengiriman_supply s ON s.po_nomor = t.po_nomor
        INNER JOIN data_supplier p ON s.idsupplier = p.id_supplier
        INNER JOIN data_barang p1 ON s.idbarang = p1.idbarang
        GROUP BY t.po_nomor
        ORDER by t.po_nomor DESC;");
      $no = 0;
      while($data = mysqli_fetch_array($query)){
       $no++;
       ?>
       <tr>
        <td><?= $no ?></td>
        <td><?= $data['po_nomor'] ?></td>
        <td><?= $data['nama_supplier'] ?></td>
        <td><?= $data['nama_barang'] ?></td>
        <td><?= $data['total'] ?></td>

      </tr>
      <?php
    }
    ?>   </tbody>

在这里我可以更改代码。

最佳答案

您尝试将其粘贴到您的sql命令中，如果它不起作用，则说明她内部有问题。

SELECT t.po_nomor, p.nama_supplier, p1.nama_barang , t.total FROM
        (SELECT t1.po_nomor, (SUM(t1.jumlah)-SUM(COALESCE(t4.terima,0))over(PARTITION by t4.refrence) ) as total
        FROM pengiriman_supply t1
        INNER JOIN data_supplier t2 ON t1.idsupplier = t2.id_supplier
        INNER JOIN data_barang t3 ON t1.idbarang = t3.idbarang
        LEFT JOIN masuk t4 ON t4.refrence = t1.po_nomor
        where t1.tanggal BETWEEN date_sub(curdate(), interval 120 day) AND curdate()
        group by t1.po_nomor,t4.po_nomor) t
        INNER JOIN pengiriman_supply s ON s.po_nomor = t.po_nomor
        INNER JOIN data_supplier p ON s.idsupplier = p.id_supplier
        INNER JOIN data_barang p1 ON s.idbarang = p1.idbarang
        GROUP BY t.po_nomor
        ORDER by t.po_nomor DESC;

关于php - mysqli_fetch_array()中的警告，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59297406/