本文介绍了警告:mysqli_num_rows()期望参数1为mysqli_result的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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