连接2个表并按ID显示


Join 2 tables and display by ID

我有两个表:

表clienti:

CodClient | NumeClient | AdresaClient 
_____________________________________
     1    |   cosmin   |   sibiu        
     2    |    alex    |   brasov

表facturi:

CodClient | NrFactura  | DataFactura
_____________________________________
     1    |   654321   | 2015-02-21        
     2    |   123456   | 2015-02-22  

CodClient=两个表之间的链接。

我有一个页面,其中有一个包含上表信息的表格:

 NumeClient | AdresaClient | NrFactura | DataFactura
  cosmin    |    sibiu     |   654321  | 2015-02-21 | Open separate
   alex     |    brasov    |   123456  | 2015-02-22 | Open separate

当我点击"打开单独"按钮时,它会将我带到这个页面,在那里我可以单独看到每一行打开单独的按钮将CodClient发送到此页面:

<?php
include('connect.php');
$CodClient=$_GET['CodClient'];
$CodClient = mysqli_real_escape_string($con,$CodClient);

$rez=mysqli_query($con,"SELECT NumeClient, AdresaClient, NrFactura, DataFactura, FacturaRetea FROM clienti c join facturi f on c.CodClient = f.CodClient");
$rows=mysqli_fetch_array($rez);
?>

  <div class="factura_header">
    <img src="img/logo2.png">
  </div>
  <div class="factura_stanga">
    <p>Client: <?php echo $rows['NumeClient'];?></p>
    <p>Adresa: <?php echo $rows['AdresaClient'];?></p>
  </div>
  <div class="factura_dreapta">
    <p>Numar factura: <?php echo $rows['NrFactura'];?></p>
    <p>Data Factura: <?php echo $rows['DataFactura'];?></p>
  </div>
  <div class="despartitor">
  </div>
</div>
<?php
mysqli_close($con);
?>

问题是,无论我打开哪一行,它都会向我显示相同的信息。如果我单击第一个"打开分隔",它会显示带有"cosmin-sibiu"的信息,如果我单击第二个"打开隔离",它也会显示"cosmin-sbiu",尽管在浏览器地址栏中我看到了corect-CodClient。我做错了什么?抱歉我英语不好。

在您的SQL中,您没有传递$CodClient id,而是简单地根据属性之间的一般关系连接两个SQL表。。。您需要添加类似SELECT NumeClient, AdresaClient, NrFactura, DataFactura, FacturaRetea FROM clienti c join facturi f on c.CodClient = f.CodClient where c.CodClient = 10 /* this is your $CodClient id from URL */的内容。。。显然要确保它不受注射等的影响。