警告：mysqli_num_rows（）期望参数1为mysqli_result

问题描述

 elseif（isset（$ _ POST ['  search'] ））{
 
＃单击保存按钮
  $ vname  = $ _POST ['  vendorname']; 
 
  //  连接到mysql  
  $ link  = mysqli_connect（  localhost，  root， ，  magna-billing）; 
  //   mysql搜索查询 
  $ query  =   SELECT`vendoraddress`，`vendorcontact`，'vendorgst'，'vendorstateid'， 'vendorcode'FROM'defalldetail` WHERE`vendorname` = $ vname; 
  $ result  = mysqli_query（$ link，$ query）; 
 if（mysqli_num_rows（$ result）> 0）
 {
  while （$ row = mysqli_fetch_array（$ result））
 {
 
  $ vadd  = $ row ['  vendoraddress' ]; 
  $ vcont  = $ row ['  vendorcontact ]; 
  $ vgst  = $ row ['  vendorgst ]; 
  $ vstate  = $ row ['  vendorstateid ]; 
  $ vcode  = $ row ['  vendorcode ]; 
 
} 
 
} 
 
 mysqli_free_result（$ result）; 
 mysqli_close（$ link）; 
 
  else  {
  echo   未定义; 
 
} 
 
} 
 
  else  {
  $ vadd  =  ; 
  $ vcont  =  ; 
  $ vgst  =  ; 
  $ vstate  =  ; 
  $ vcode  =  ; 
} 
 
 
 
 
 ？>  

我尝试了什么：



我试图显示数据库单击搜索按钮后文本框中的值，这里我卡在某处。获得警告：mysqli_num_rows（）期望参数1为mysqli_result错误。在此先感谢。

解决方案

elseif (isset($_POST['search'])) {

        # Save-button was clicked
        $vname = $_POST['vendorname'];

        // connect to mysql
        $link =  mysqli_connect("localhost", "root", "", "magna-billing");
        // mysql search query
        $query = "SELECT  `vendoraddress`, `vendorcontact`, 'vendorgst', 'vendorstateid', 'vendorcode' FROM `vendordetail` WHERE `vendorname` = $vname ";
        $result = mysqli_query($link, $query) ;
        if(mysqli_num_rows($result) > 0)
        {
        while ($row = mysqli_fetch_array($result))
        {

          $vadd = $row['vendoraddress'];
          $vcont = $row['vendorcontact'];
          $vgst = $row['vendorgst'];
          $vstate = $row['vendorstateid'];
          $vcode = $row['vendorcode'];

}

}

mysqli_free_result($result);
mysqli_close($link);

else {
  echo "Undefined";

}

}

 else {
  $vadd = "";
  $vcont = "";
  $vgst = "";
  $vstate = "";
  $vcode = "";
}




?>

What I have tried:

I am trying to show the database value in the text box after clicking on search button, Here I stuck somewhere. Getting Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result error. Thanks in advance.

解决方案

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