2017-05-29 16 views
1

映像路徑上不顯示爲文本框

<script type="text/javascript"> 
 

 
function myFunction() { 
 
//var query = '?'; 
 
    var str = $("form").serialize(); 
 
    str = str.replace(/\=/g, "='").replace(/\&/g, "'&"); 
 
    $("#results").text(str+"'"); 
 
    var x = $("#results").text(); 
 
return x; 
 

 
} 
 
    
 
$('#myButton').on('click',function(){  
 
    var jsonString = JSON.stringify(myFunction()); 
 
    console.log(jsonString); 
 

 
$.ajax({ 
 
    url: 'insert_value.php', 
 
    data: jsonString, 
 
    contentType: 'application/json; charset=utf-8', 
 
    type: 'POST', 
 
    }).done(function(resp) { 
 
    $('#result').html(resp) 
 

 
    }); 
 
    }); 
 

 
    </script>
<!DOCTYPE html> 
 
<html> 
 
<head> 
 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
    <script src="https://code.jquery.com/jquery-1.10.2.js"></script> 
 
    <link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css" rel="stylesheet" /> 
 

 
</head> 
 
<body> 
 

 

 
<?php 
 
include("con_gen.php"); 
 

 
error_reporting(E_ALL); 
 

 
// DATABASE CONNECTION AND SELECTION VARIABLES - GET THESE FROM YOUR HOSTING COMPANY 
 
$db_host = "localhost"; // PROBABLY THIS IS OK 
 
$db_name = "idcard"; 
 
$db_user = "root"; 
 
$db_word = ""; 
 

 
// OPEN A CONNECTION TO THE DATA BASE SERVER AND SELECT THE DB 
 
$mysqli = new mysqli($db_host, $db_user, $db_word, $db_name); 
 

 
// DID THE CONNECT/SELECT WORK OR FAIL? 
 
if ($mysqli->connect_errno) 
 
{ 
 
    $err 
 
    = "CONNECT FAIL: " 
 
    . $mysqli->connect_errno 
 
    . ' ' 
 
    . $mysqli->connect_error 
 
    ; 
 
    trigger_error($err, E_USER_ERROR); 
 
} 
 

 
// RUN A QUERY 
 
$result = mysqli_query($mysqli,"SELECT value FROM combo1"); 
 
$num_rows = mysqli_num_rows($result); 
 

 
//echo "$num_rows Rows\n"; 
 
if ($result->num_rows > 0) { 
 
    // output data of each row 
 
    $array = Array(); 
 
    $array1 = Array(); 
 
    while($row = $result->fetch_assoc()) { 
 
     //echo "<br> value: ". $row['value']. "<br>"; 
 
      $array[] = $row['value']; 
 
     
 
} 
 
//print_r($array); 
 

 
$sql = "SELECT static_name FROM static_values"; 
 
$result = mysqli_query($mysqli, $sql); 
 
if ($result && mysqli_num_rows($result) > 0) 
 
while($row = mysqli_fetch_array($result)){ 
 
       //echo "<option>" . $row['static_name'] . "</option>"; 
 
       $array1[]=$row['static_name']; 
 
        
 
       //echo $source; 
 
       } 
 
       // print_r($array1); 
 
       foreach ($array as $row) 
 
{ 
 
    if(in_array($row, $array1)) 
 
    { 
 
$sql = "SELECT source_table,Alias_name FROM static_values where static_name='$row'"; 
 
$res = $mysqli->query($sql); 
 

 
// DID THE QUERY WORK OR FAIL? 
 
if (!$res) 
 
{ 
 
    $err 
 
    = 'QUERY FAILURE:' 
 
    . ' ERRNO: ' 
 
    . $mysqli->errno 
 
    . ' ERROR: ' 
 
    . $mysqli->error 
 
    . ' QUERY: ' 
 
    . $sql 
 
    ; 
 
    trigger_error($err, E_USER_ERROR); 
 
} 
 

 
// ARE THERE ANY ROWS IN THE RESULTS SET 
 

 
if ($res->num_rows == 0) 
 
{ 
 
    trigger_error("ERROR: NO DATA FOUND BY $sql", E_USER_ERROR); 
 
} 
 

 
// RETRIEVE THE ROWS INTO AN ARRAY OF HTML STATEMENTS 
 
//$html = ""; 
 
echo '<form method="POST" action="" id="myform">'; 
 

 
while ($row = $res->fetch_object()) 
 
{ 
 
    //$html =$html.' 
 
    echo' 
 
<tr> 
 
    <td> 
 
    <div class="row-fluid"> 
 

 
     <div class="span3 bgcolor"> 
 
    <label>'.$row->Alias_name.'</label> 
 

 

 
     <select id='.$row->source_table.' name='.$row->source_table.' data-live-search="true" class="selectpicker form-control" type="sel">'; 
 

 
    /*foreach ($Data->{$row->source_table} as $key =>$item) { 
 
    echo "<option value=".$key.">".$item."</option>" ; 
 
    }*/ 
 
    foreach ($Data->{$row->source_table} as $key =>$item) { 
 
    echo "<option value=".$key.">".$item."</option>"; 
 
    } 
 
echo '   
 
      </select> 
 
              </div> 
 
     </div> 
 
    </td> 
 
</tr>'; 
 
} 
 
} 
 
else if($row=="Image") 
 
{ 
 
    echo ' 
 
<tr> 
 
    <td> 
 
    <div class="row-fluid"> 
 
       
 
      <div class="span3 bgcolor"> 
 
      <label>'.$row.'</label> 
 
       <label>'.$row.'</label> 
 
       <input type="file" name="'.$row.'" id="'.$row.'" onchange="document.getElementById(\''.$row.'\').value = this.value"> 
 
       <input type="text" name="'.$row.'" id="'.$row.'" value="'.$imagepath.'"> 
 
      
 
</div> 
 
</div> 
 
    </td> 
 
</tr>'; 
 
    } 
 
    else 
 
    { 
 
    //$html =$html.' 
 
    echo ' 
 
<tr> 
 
    <td> 
 
    <div class="row-fluid"> 
 
       
 
      <div class="span3 bgcolor"> 
 
      <label>'.$row.'</label> 
 
       <input id='.$row.' type="text" placeholder=" Enter Value " name='.$row.' style="width:100%" class="form-control" required="" /> 
 
           
 
       </select> 
 
</div> 
 
</div> 
 
    </td> 
 
</tr>'; 
 

 
} 
 

 
} 
 
echo'<input id="myButton" type="button" value="SUBMIT" class="btn btn-success" name="submit" onclick="myFunction();"/>'; 
 
echo '<p id="results">&nbsp;</p>'; 
 
echo '<pre id="result"></pre></form> '; 
 

 
//echo '<p><tt id="results"></tt></p>'; 
 
} 
 

 

 

 
       ?> 
 
       </body> 
 
</html>

<?php else if($row=="Image") 
 
{ 
 
    echo '<tr> 
 
      <td> 
 
       <div class="row-fluid"> 
 
        <div class="span3 bgcolor"> 
 
         <label>'.$row.'</label> 
 
\t \t \t  <input type="file" name='.$row.' id='.$row.' onchange="document.getElementById('.$row.').value = this.value"> 
 
\t \t \t \t <input type="text" name='.$row.' id='.$row.'> 
 
\t \t  </div> 
 
\t \t </div> 
 
\t  </td> 
 
\t </tr>'; 
 
?>

在這裏,我讀文本內的圖像完整路徑和插入到數據庫中,但這裏的圖像路徑沒有進入文本框請告訴我解決方案

我發佈了我的整個代碼,請讓我知道soluti關於如何將圖像路徑導入文本框

+0

你應該使用'value'屬性在文本框中 –

+0

檢查這個https://stackoverflow.com/questions/5812274/auto-populating-a-textarea-with-php-variables –

+0

你能張貼休息這段代碼之前和之後的代碼?這會讓你更容易理解你想要達到的目標。 –

回答

0

你需要價值屬性,每個屬性值應該用單引號或雙引號這樣

<?php 
echo '<tr> 
     <td> 
      <div class="row-fluid"> 
       <div class="span3 bgcolor"> 
        <label>'.$row.'</label> 
        <input type="file" name="'.$row.'" id="'.$row.'" onchange="document.getElementById(\''.$row.'\').value = this.value"> 
        <input type="text" name="'.$row.'" id="'.$row.'" value="'.$imagepath.'"> 
       </div> 
      </div> 
     </td> 
    </tr>'; 
?> 
+0

ya試着給出這樣的提示:未定義變量:imagepath在C:\ xampp \ htdocs \ dyn_html.php on line 136 – shruti

+0

@shruti你需要自己輸入'$ imagepath'的值 – Akintunde007

+0

對不起,我收到你要告訴的內容 – shruti

0

檢查此示例代碼並驗證您獲取的ID是否正確。

<!DOCTYPE html> 
 
<html> 
 
<head> 
 
</head> 
 
<body> 
 
<div class="data"> 
 
<input type="file" name='fileTest' id='FileTest' onchange="document.getElementById('test').value = this.value"> 
 
    <input type="text" name='test' id='test'> 
 
</div> 
 
</body> 
 
</html>

+0

上面的代碼會給出錯誤對我來說 – shruti

+0

嘗試檢查頁面呈現後的id名稱是否嵌入在雙重qoutes中 –

+0

你可以在瀏覽器頁面呈現後更新代碼 –

0

好了,所以看代碼和您的意見一個封閉看起來$imagepath在被調用之前沒有被聲明。您需要聲明$imagepath,然後在調用它之前設置它的值。在調用變量之前未聲明該變量將導致錯誤,如果未設置變量,將導致在調用變量時不返回任何內容。