陣推未定義

Question

我嘗試使用PHP

$select_all_schools = "SELECT * FROM schools "; 
$query = mysql_query($select_all_schools) OR die(mysql_error()); 

while($row = mysql_fetch_array($query)){ 
    $item=array( 
     'school_name' => $row['school_name'], 
     'school_address' => $row['school_address'], 
     'school_id' => $row['school_id'], 
    ); 

    $select_sections = "SELECT * FROM section WHERE school_id = '".$row['school_id']."'"; 

    $query_section = mysql_query($select_sections) or die(mysql_error()); 
    $sections_counts = mysql_num_rows($query_section); 
    $select_sections_deped_archive = "SELECT * FROM deped_grade_archive 
             WHERE school_id = '".$row['school_id']."' 
             GROUP BY section_id "; 

    $query_section_deped_archive = mysql_query($select_sections_deped_archive) or die(mysql_error()); 
    $sections_counts_grade_archive = mysql_num_rows($query_section_deped_archive); 

    if($sections_counts_grade_archive == $sections_counts){ 
     $item['stat'] = 'Complete'; 
    } 
    else{ 
     $item['stat'] ='Incomplete'; 
    } 
}   

echo json_encode($item); 

然後在我的數組推新值用ajax

function get_all_school_status(){ 
    $.ajax({ 
     type:'POST', 
     url:'deped_functions.php', 
     dataType:'json', 
     data:{'func_num':'1'}, 
     success:function (data){ 
      $.each(data, function(i, item) { 
       html = "<tr>"; 
       html += "<td style='width:20%;'><input type='radio' name='school_id' value='"+data[i].school_id+"'></td>"; 
       html += "<td style='width:25%;'><label>"+data[i].stat+"</label></td>"; 
       html += "<td style='width:55%;'><label >"+data[i].school_name+"</label></td>"; 
       html += "</tr>"; 

       $('#table-schools-content').append(html); 
      }); 
     } 
    }); 
} 
get_all_school_status(); 

但不幸的是即時得到未定義的值，雖然得到的值我控制檯顯示我正確地從php到ajax的值。我做了什麼錯事？請幫助傢伙。 tnx

Answer 1

Js部分代碼是正確的，但是您的php數組形成了錯誤。
嘗試替換上面的代碼。錯誤評論。

$count = 0; // keys of array 
while($row = mysql_fetch_array($query)){ 
    $count++; // in prevoius version you get only one row from table 
    // correct version of school array 
    $item[$count] = array( 
     'school_name' => $row['school_name'], 
     'school_address' => $row['school_address'], 
     'school_id' => $row['school_id'], 
    ); 

    $select_sections = "SELECT * FROM section WHERE school_id = '".$row['school_id']."'"; 

    $query_section = mysql_query($select_sections) or die(mysql_error()); 
    $sections_counts = mysql_num_rows($query_section); 
    $select_sections_deped_archive = "SELECT * FROM deped_grade_archive 
             WHERE school_id = '".$row['school_id']."' 
             GROUP BY section_id "; 

    $query_section_deped_archive = mysql_query($select_sections_deped_archive) or die(mysql_error()); 
    $sections_counts_grade_archive = mysql_num_rows($query_section_deped_archive); 

    if($sections_counts_grade_archive == $sections_counts){ 
     // success or error message with key 'status' 
     $item[$count]['status'] = 'Complete'; 
    } 
    else{ 
     $item[$count]['status'] = 'Incomplete'; 
    } 
} 

這樣做的結果是： http://clip2net.com/s/2MJmt

Answer 2

jQuery .each method不是用來遍歷數組，而是用HTML元素。你的學校數組應該進行遍歷正常使用FOR循環：

success: function(data) { 
       if(typeof data == "string") //handle PHP errors - this happens if non-json string is returned by a server 
        alert("JQuery failed to convert data to JS object!"); 
       for(var i=0; i<data.length; i++) { 
        /*Do whatever with data[i]*/ 
       } 
     } 

如果數據數組包含類似數據的非*命令鍵[「school1」]或類似的東西，你必須使用for(var i in data)它類似於PHP的foreach 。

此外，我不確定，但我認爲你應該有dataType:'application/json'，這是正確的json mime type。