我该如何修复此警告:mysqli_fetch_array()期望参数1为mysqli_result,给定布尔值 [英] How can I fix this warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given

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问题描述

请某人帮助我。我很紧急。我无法解决这个问题。





这是我的结果表格



<?php

session_start();

?>







< meta charset =utf-8>

< meta name =viewportcontent =width = device-width,initial-scale = 1>

























< h3 class =panel-title> Laporan Bayaran















Nama Koperasi:





--Nama Koperasi--

<?php

//第一个cari无法连接ke数据库

include_once(./includes / dbconnection.php);



$ sql =SELECT * FROM coop01;

$ result = mysqli_query($ conn,$ sql);



while($ coop = mysqli_fetch_array($ result)){

$ c_id = $ coop ['coopId'];

$ c_name = $ coop ['coopName'];



echo

。$ c_name。



;

}



?>










Jenis Carian:


< br>

--Sila Pilih--

Resit

Baucer

Invois Jualan

Invois Belian












Tarikh:





























































/ *样式搜索字段* /

form.example输入[type = text] {

padding:8px;

font-size:15px;

border:1px solid grey;

float:left;

宽度:80%;

背景:#f1f1f1;

}



/ *样式提交按钮* /

form.example按钮{

float:left;

宽度:20%;

填充:10px;

背景:#2196F3;

颜色:白色;

字体大小:17px;

border:1px solid grey;

border-left:none; / *防止双边框* /

光标:指针;

}



form.example按钮:悬停{

背景:#0b7dda;

}



/ * Clear floats * /

form.example :: after {

content:;

clear:both;

display:table;

}



我尝试过:



这是我提交按钮搜索后的结果











< meta charset =utf-8>

< meta name =viewportcontent =width = device-width,initial-scale = 1> ;











sql =SELECT * FROM coop01;


result = mysqli_query(


conn,


please someone help me . im in urgent situation. i cant solved this.


this is my result form

<?php
session_start();
?>



<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">












Laporan Bayaran









Nama Koperasi :



--Nama Koperasi--
<?php
//First cari fail untuk connection ke database
include_once ("./includes/dbconnection.php");

$sql = "SELECT * FROM coop01";
$result = mysqli_query($conn, $sql);

while ($coop = mysqli_fetch_array($result)) {
$c_id = $coop['coopId'];
$c_name = $coop['coopName'];

echo "
".$c_name."

";
}

?>





Jenis Carian :



--Sila Pilih--
Resit
Baucer
Invois Jualan
Invois Belian






Tarikh:
































/* Style the search field */
form.example input[type=text] {
padding: 8px;
font-size: 15px;
border: 1px solid grey;
float: left;
width: 80%;
background: #f1f1f1;
}

/* Style the submit button */
form.example button {
float: left;
width: 20%;
padding: 10px;
background: #2196F3;
color: white;
font-size: 17px;
border: 1px solid grey;
border-left: none; /* Prevent double borders */
cursor: pointer;
}

form.example button:hover {
background: #0b7dda;
}

/* Clear floats */
form.example::after {
content: "";
clear: both;
display: table;
}

What I have tried:

this i my outcome after submit button search





<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">





解决方案

sql = "SELECT * FROM coop01";


result = mysqli_query(


conn,


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