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Php 显示数据库中的值列表 $sql=“从主题中选择*”; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $id=$row[“id”]; $name=$row[“name”]; $developers=$row[“developers”]; $description=$row[“description”]; $category=$row[“category”]; $logo=$row[“logo”]; $officialLink=$row[“link”]; ////////////////////// $sql=“从文件中选择*,其中软件_id='$id'; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $version=$row[“version”]; $file=$row[“file”]; $link=$row[“link”]; $created=$row[“created”]; $versions.=''; $logos=“”; } ////////////////////// $solfList.=' “.$name。” '; }_Php_Mysqli - Fatal编程技术网

Php 显示数据库中的值列表 $sql=“从主题中选择*”; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $id=$row[“id”]; $name=$row[“name”]; $developers=$row[“developers”]; $description=$row[“description”]; $category=$row[“category”]; $logo=$row[“logo”]; $officialLink=$row[“link”]; ////////////////////// $sql=“从文件中选择*,其中软件_id='$id'; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $version=$row[“version”]; $file=$row[“file”]; $link=$row[“link”]; $created=$row[“created”]; $versions.=''; $logos=“”; } ////////////////////// $solfList.=' “.$name。” '; }

Php 显示数据库中的值列表 $sql=“从主题中选择*”; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $id=$row[“id”]; $name=$row[“name”]; $developers=$row[“developers”]; $description=$row[“description”]; $category=$row[“category”]; $logo=$row[“logo”]; $officialLink=$row[“link”]; ////////////////////// $sql=“从文件中选择*,其中软件_id='$id'; $query=mysqli\u query($db\u conx,$sql); while($row=mysqli\u fetch\u数组($query,mysqli\u ASSOC)){ $version=$row[“version”]; $file=$row[“file”]; $link=$row[“link”]; $created=$row[“created”]; $versions.=''; $logos=“”; } ////////////////////// $solfList.=' “.$name。” '; },php,mysqli,Php,Mysqli,此代码仅显示一个值。我想显示表中存储的所有值。我有这个密码。但是我不明白我在哪里犯了错误。就像我说的,运行内部查询会丢失所有外部查询结果集。因此,您的代码应该如下所示: $sql = "SELECT * FROM themes"; $query = mysqli_query($db_conx, $sql); while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) { $id = $row["id"];

此代码仅显示一个值。我想显示表中存储的所有值。我有这个密码。但是我不明白我在哪里犯了错误。

就像我说的,运行内部查询会丢失所有外部查询结果集。因此,您的代码应该如下所示:

$sql   = "SELECT * FROM themes";
$query = mysqli_query($db_conx, $sql);

while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
    $id           = $row["id"];
    $name         = $row["name"];
    $developers   = $row["developers"];
    $description  = $row["description"];
    $category     = $row["category"];
    $logo         = $row["logo"];
    $officialLink = $row["link"];

    //////////////////////

    $sql          = "SELECT * FROM file WHERE software_id = '$id'";
    $query        = mysqli_query($db_conx, $sql);
    while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
        $version = $row["version"];
        $file    = $row["file"];
        $link    = $row["link"];
        $created = $row["created"];

        $versions .= '<li><a href="../files/' . $file . '"                                target="_blank">
<i class="fa fa-mail-forward">
</i> ' . $version . '</a></li>';
        $logos = "<img src ='../img/logos/{$logo}' >";
    }

    //////////////////////

    $solfList .= '
                <li>
                <h3>' . $name . '</h3>

                </li>
                ';

}

由于在
sql
中使用相同的变量,您将丢失第一个循环的值。因此,对两个查询及其结果使用两个不同的变量

$sql = "SELECT * FROM themes";
$query = mysqli_query($db_conx, $sql); 
while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)){ 
$id = $row["id"];

// your code

$sql = "SELECT * FROM file WHERE software_id = '$id'";
$result = mysqli_query($db_conx, $sql); 
    while ($r = mysqli_fetch_array($result, MYSQLI_ASSOC)){ 
        $version = $r["version"];

        // your code

    }

    // your code

}
$sql=“从主题中选择*”;
$query=mysqli\u query($db\u conx,$sql);
而($row_outer=mysqli_fetch_数组($query,mysqli_ASSOC)){
$id=$row_outer[“id”];
$name=$row_outer[“name”];
$developers=$row_outer[“developers”];
$description=$row_outer[“description”];
$category=$row_outer[“category”];
$logo=$row_outer[“logo”];
$officialLink=$row_outer[“link”];
//////////////////////
$solfList=“”;
$sql\u internal=“从文件中选择*,其中软件\u id='$id';
$query\u inner=mysqli\u query($db\u conx,$sql\u inner);
而($row_inner=mysqli_fetch_array($query_inner,mysqli_ASSOC)){
$version=$row_inner[“version”];
$file=$row_inner[“file”];
$link=$row_inner[“link”];
$created=$row_inner[“created”];
$versions.='
  • '; $logos=“”; } ////////////////////// $solfList.='
  • “.$name。”
  • '; }
    查看这两条查询语句。运行内部查询将丢失所有外部查询结果集;在while循环之前。。当我想象你可能还想要
    $logos[]=…
    $logos.=…
    @urfusion时,这不是变量
    $sql
    ,而是外部结果集被内部结果集破坏。变量
    $sql
    可以再次使用。
    $sql = "SELECT * FROM themes";
    $query = mysqli_query($db_conx, $sql); 
                while ($row_outer = mysqli_fetch_array($query, MYSQLI_ASSOC)){ 
                $id = $row_outer["id"];
                $name = $row_outer["name"];
                $developers = $row_outer["developers"];
                $description = $row_outer["description"];
                $category = $row_outer["category"];
                $logo = $row_outer["logo"];
                $officialLink = $row_outer["link"];
                //////////////////////
                $solfList = "";
                $sql_inner = "SELECT * FROM file WHERE software_id = '$id'";
                $query_inner = mysqli_query($db_conx, $sql_inner); 
                    while ($row_inner = mysqli_fetch_array($query_inner, MYSQLI_ASSOC)){ 
                    $version = $row_inner["version"];
                    $file = $row_inner["file"];
                    $link = $row_inner["link"];
                    $created = $row_inner["created"];
                        $versions .='<li><a href="../files/' .$file. '"                                target="_blank">
    <i class="fa fa-mail-forward">
    </i> ' .$version. '</a></li>';
                        $logos = "<img src ='../img/logos/{$logo}' >";
                }
                        //////////////////////
                    $solfList .='
                    <li>
                    <h3>' .$name. '</h3>
    
                    </li>
                    ';
    
                }