php - 使用GET方法并在mysqli_fetch_array（）上获得警告

It seems that my code is wrong. I'm using method GET and pass the variable on its page. This is my form code.

看来我的代码错了。我正在使用方法GET并在其页面上传递变量。这是我的表单代码。

<form method="get" action="?nama'">
    <div class="input-group">
        <span class="input-group-addon" id="basic-addon1"><i class="fa fa-search"></i></span>
        <input type="text" class="form-control" name="nama" placeholder="Cari berdasarkan Nama Lengkap Pendaftar" aria-describedby="basic-addon1">
        <span class="input-group-btn">
            <button class="btn btn-success" type="submit">Go!</button>
        </span>
    </div>
</form>

And this is for the GET method. I'm using table to write the result.

这是针对GET方法的。我正在使用表来写结果。

<div class="table-responsive">      
    <table class="table table-bordered table-hover">
        <tr style="font-weight: bold; text-align: center;">
            <td> No. </td>
            <td> ID Pendaftaran </td>
            <td> Nama Lengkap </td>
            <td> Jenis Kelamin </td>
            <td> Asal Sekolah </td>
            <td> Tanggal Daftar </td>
            <td> Pengembalian Form </td>                            
        </tr>
        <?php
        if (isset($_GET['nama'])) {
            $no = 1;
            $nama = $_GET['nama'];
            $query = mysqli_query($koneksi, "SELECT * FROM pendaftar WHERE NAMAPENDAFTAR LIKE '%".$nama."'%'");
            while ($loop = mysqli_fetch_array($query)) {
        ?>
        <tr style="text-align: center;">
            <td> <?php echo $no; ?> </td>
            <td> <?php echo $loop['IDPENDAFTAR']; ?> </td>
            <td> <?php echo $loop['NAMAPENDAFTAR']; ?> </td>
            <td> <?php echo $loop['JENISKELAMIN']; ?> </td>
            <td><?php echo $loop['SEKOLAHASAL']; ?> </td>
            <td><?php echo date('d-m-Y', strtotime($loop['TANGGALDAFTAR'])); ?> </td>
            <td><button type="button" name="kembali" class="btn btn-default" data-toggle="modal" data-target="#myModal">Form Kembali</button></td>
        </tr>
        <?php
            $no++;
            }
        }
        else {
            $no = 1; 
            $query1 = mysqli_query($koneksi, "SELECT * FROM pendaftar");
            while ($loop = mysqli_fetch_array($query1)) {
        ?>
        <tr style="text-align: center;">
            <td> <?php echo $no; ?> </td>
            <td> <?php echo $loop['IDPENDAFTAR']; ?> </td>
            <td> <?php echo $loop['NAMAPENDAFTAR']; ?> </td>
            <td> <?php echo $loop['JENISKELAMIN']; ?> </td>
            <td><?php echo $loop['SEKOLAHASAL']; ?> </td>
            <td><?php echo date('d-m-Y', strtotime($loop['TANGGALDAFTAR'])); ?> </td>
            <td><button type="button" name="kembali" class="btn btn-default" data-toggle="modal" data-target="#myModal">Form Kembali</button></td>
        </tr>
        <?php
            $no++;
            }
        }
        ?>
    </table>
</div>

It gets error like this.

它会像这样出错。

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\WebKP\admin-site.php on line 126 But while I'm trying to query on my phpmyadmin, it has the results.

警告：mysqli_fetch_array（）期望参数1为mysqli_result，第126行的C：\ xampp \ htdocs \ WebKP \ admin-site.php中给出布尔值但是当我尝试查询我的phpmyadmin时，它有结果。

1 个解决方案

#1

You have SQL syntax error here

这里有SQL语法错误

$query = mysqli_query($koneksi, "SELECT * FROM pendaftar WHERE NAMAPENDAFTAR LIKE '%".$nama."'%'");

Use below code

使用下面的代码

 $query = mysqli_query($koneksi, "SELECT * FROM pendaftar WHERE NAMAPENDAFTAR LIKE '%".$nama."%'")

#1