PHP-使用字段名称作为变量[重复]

Question

我得到的错误"警告：mysql_field_name（）期望参数1是资源，对象在...在第28行"

我对PHP相当陌生，但我试图完成的是读取超文本标记语言文件，并用记录的值替换自定义标记。标签结构是|FIELD-NAME-Recage#|例如，如果我的sql返回两个记录的"名称"字段，它将在超文本标记语言文件中查找以下两个标签|Name1|和|Name2|，并用返回的两个替换它价值观。说亚当和吉姆。

下面是我到目前为止的代码。

$c = '|+Name1+|<br>|+Name2+|';

echo(ReplaceHTMLVariables(mysqli_query($con,"SELECT * FROM nc_names"),$c));

function ReplaceHTMLVariables($results, $content){
    while($row = mysqli_fetch_array($results)){
        //Define a variable to count what record we are on
        $rNum = 1;

        //Loop through all fields in the record
        $field = count( $row );    
        for ( $i = 0; $i < $field; $i++ ) {
            //Use the field name and record number to create variables to look for then replace with the current record value
            $content = str_replace("|+".mysql_field_name( $results, $i ).$rNum."+|",$row[$i],$content);
        }

        //move to next record
        $rNum++;
    }
    return $content;
}

第28行引用了这一行

$content=str_replace（“|”）.mysql_field_name（$results，$i）$纳姆。“|”，$行[$i]，$内容）；

Answer 1

@Barmar的评论是正确的，您不能混合使用mysql_u和mysqli_u函数，这就是为什么您要声明错误的原因

我还对你的代码做了一些其他的修改来简化它。有关解释，请参阅内联注释

$c = '|+Name1+|<br>|+Name2+|';

// database query on separate line, not in function call, so we can catch any errors
$res = mysqli_query($con,"SELECT * FROM nc_names") or die(mysqli_error());

echo(ReplaceHTMLVariables($res,$c));


function ReplaceHTMLVariables($results, $content){

    //Define a variable to count what record we are on
    $rNum = 1;

    // use fetch_assoc instead of fetch_array
    // fetch_assoc will give you an array of column_name => value pairs
    while($row = mysqli_fetch_assoc($results)){

        // MOVED this outside the loop or you will reset to 1 it for each loop through
        //Define a variable to count what record we are on
        //$rNum = 1;

        // extract column names from associative array
        $field_names = array_keys($row);

        //Loop through all fields in the record

        // use foreach loop instead of for loop (this is just my preference, not essential)            // $field = count( $row );    
        //for ( $i = 0; $i < $field; $i++ ) {
        foreach ($field_names as $field_name) { 
            // Use the field name and record number to create variables to look for then replace with the current record value

            // build placeholder on separate line, makes code easier to read
            $placeholder = "|+{$field_name}{$rNum}+|"; 

            $content = str_replace($placeholder, $row[$field_name], $content);
        }

        // move to next record
        $rNum++;
    }

    return $content;
}

Answer 2

你正在混合MySQL和MySQLi，如果你拉mysqli_fetch_assoc（），你真的不需要做所有这些：

function ReplaceHTMLVariables($results, $content)
{
    //Define a variable to count what record we are on
    $rNum = 1;

    /*** Using Mysqli::fetch_assoc() ***/
    while( $row = mysqli_fetch_assoc( $results ) )
    {
        //Loop through all fields in the record
        /*** Variable $value passed by reference for performance ***/
        foreach( $row as $fieldName => &$value ) {
            $content = str_replace("|+".$fieldName.$rNum."+|",$value,$content);
        }

        ++$rNum; /*** Faster than $rNum++ **/
    }
    return $content;
}

mysqli_fetch_assoc（）将数据作为关联数组，字段名作为索引键。