Question

我的一个阵列出了问题。出于某种原因，每当我测试它时，我都会收到此错误：

Unknown column 'ext_token' in 'where clause'如果我有错误的mysql错误，
或者我得到Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given，如果我拿出来的话。

我不确定它是否与数组中第一个columnName为ext_token的事实有关，但我认为我会通过if if去除它。

以下是数组的代码：

   $mat_total = array();

foreach ($result_array1 as $columnName => $columnData){

   if($columnName != "ext_token" || "ext_token_child"){
           var_dump($columnName);
           $mat_sql = mysql_query("SELECT * FROM materials WHERE mat_token = $columnName");
           $mat_array = mysql_fetch_assoc($mat_sql);
           $material_tok = $mat_array['mat_token'];
           $material_price_unit = $mat_array['material_price_per_unit'];


           $total_mat_price = $material_price_unit * $columnData;

           array_push($mat_total, "$material_tok => $total_mat_price");

   }
   else{
       echo "hello";    
   }

}

感谢您的帮助。

编辑：就数组来说，这是一个更清晰的版本

Column name: ext_token Column data: roof
Column name: ext_token_child Column data: felt
Column name: concrete Column data: 4
Column name: cement Column data: 3
Column name: sand Column data: 2
Column name: wood_4_2 Column data: 4
Column name: wood_8_2 Column data: 2
Column name: felt Column data: 2

或者有这样的：

array(8) { ["ext_token"]=> string(4) "roof" ["ext_token_child"]=> string(4) "felt" ["concrete"]=> string(1) "4" ["cement"]=> string(1) "3" ["sand"]=> string(1) "2" ["wood_4_2"]=> string(1) "4" ["wood_8_2"]=> string(1) "2" ["felt"]=> string(1) "2" }

Answer 1

这个表达似乎错了：

$columnName != "ext_token" || "ext_token_child"

我认为你的意思是

$columnName != "ext_token" && $columnName != "ext_token_child"

Answer 2

您忘记用单引号括起该值。它应该像

WHERE mat_token ='$ columnName'“

Answer 3

if($columnName != "ext_token" || "ext_token_child"){

应为if($columnName != "ext_token" && $columnName != "ext_token_child"){

或者使用==，无论你拍摄哪个。

Answer 4

试试这个：

$mat_total = array();

foreach ( $result_array1 as $columnName => $columnData ) {

  if( $columnName != "ext_token" || $columnName == "ext_token_child" ) {

    var_dump( $columnName );
    $mat_sql = mysql_query( "SELECT * FROM materials WHERE mat_token = '$columnName'" );

    $mat_array = mysql_fetch_assoc( $mat_sql );
    $material_tok = $mat_array[ 'mat_token' ];
    $material_price_unit = $mat_array[ 'material_price_per_unit' ];

    $total_mat_price = $material_price_unit * $columnData;

    array_push( $mat_total, "$material_tok => $total_mat_price" );

  } else {
    echo "hello";   
  }

}

使用数组选择获取MySQL错误

4 个答案: