2014-09-20 179 views
0

我不能得到这个工作 这是我的主页:关系数据库查询警告

<?php 

    $satt = getfn(); 
    while ($row = mysql_fetch_array($satt)){ 
    echo "<li><a href=\"\">{$row['satake']}</a></li>";} 


    $yosef = yosef($row["jid"]); 
    while ($row = mysql_fetch_array($yosef)){ 
    echo "<li>{$row['yosef']}</li>";} 



    $jidd = jid(); 
    while ($row = mysql_fetch_array($jidd)){ 
    echo "<li>{$row['jid']}</li>"; 
    } 



    ?> 

,这是我的功能:

<?php 
    /* function confirm_query($rs){ 
        if (!$rs){ 
      die("eror".mysql.error()); 
      } 
    }*/ 
function getfn(){ 
      $query = "SELECT * FROM fn";   
      $result = mysql_query($query); 
      //confirm_query($result); 
      return $result; 
} 

function yosef($row){ 
      $query = "SELECT `yosef` FROM `satake` WHERE jid = {$row}";   
      $result = mysql_query($query); 
      // confirm_query($result); 
      return $result; 
} 


function jid(){ 
      $query = "SELECT * FROM `satake` ORDER BY `satake`.`jid` ASC";   
      $result = mysql_query($query); 
       return $result; 
      // confirm_query($result); 
}   


?> 

但这给我这个错误:

警告:mysql_fetch_array()期望参数1是资源,布尔在线28 ,其中行28在这里:

$yosef = yosef($row["jid"]); 
    while ($row = mysql_fetch_array($yosef)){ 
    echo "<li>{$row['yosef']}</li>";} 

问题在哪里???

+0

谢谢你,我发现这是我的SQL查询有错误 代表的问题+ – Greatone 2014-09-20 17:54:01

回答

0

您使用$行while循环外:

while ($row = mysql_fetch_array($satt)){ 
echo "<li><a href=\"\">{$row['satake']}</a></li>";} // <-- closing bracket while 

试试这个:

while ($row = mysql_fetch_array($satt)){ 
    echo "<li><a href=\"\">{$row['satake']}</a></li>"; 


    $yosef = yosef($row["jid"]); 
    while ($row = mysql_fetch_array($yosef)){ 
      echo "<li>{$row['yosef']}</li>"; 
    } 
}