腳本來計算兩列和回聲結果

Question

我有一個工作的SQL查詢，我試圖在一個小的PHP腳本中使用，但得到分析錯誤，嘗試了很多變化。希望你能幫助。最終結果將是具有'日期'和'頻道號'的兩個字段形式，然後給出給定日期的'頻道'行數量的結果計數。對不起相當新的PHP/SQL，謝謝。

<?php 
// Connect to MSSQL and select the database 
$link = mssql_connect('localhost', 'root', '', 'jm_db'); 
mssql_select_db('jm_db'); 

// Select all our records from a table 

$mysql_query = mssql_query ('SELECT COUNT(*) FROM asterisk_cdr 
WHERE calldate LIKE '%2014-10-11%' 
    AND channel LIKE '%SIP/4546975289%'); 

echo $sql; 

?> 

我已經重新做了代碼，但得到「警告：mysql_fetch_array（）預計參數1是資源」和未定義的變量。

<?php 

// Create connection 
$mysqli = new mysqli($localhost, $root, $jm_db); 

/* check connection */ 
if (mysqli_connect_errno()) { 
printf("Connect failed: %s\n", mysqli_connect_error()); 
exit(); 
} 

$sql = ("SELECT COUNT(*) FROM asterisk_cdr 
    WHERE calldate LIKE '%2014-10-11%' 
    AND channel LIKE '%SIP/4546975289%'"); 

    $results= array(); 
while ($result = mysql_fetch_array($sql)) { 
    $results[]= $result; 
} 

foreach($results as $result){ 
    echo $result['calldate'] . " " . $result['channel']; 
} 

?> 

Answer 1

這與PDO

一個簡單的例子

<?php 
    try { 
    $dns = 'mysql:host=localhost;dbname=jm_db'; 
    $user = 'root'; 
    $pass = ''; 

    $options = array(
     PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8", 
     PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION 
    ); 
    $cnx = new PDO($dns, $user, $pass, $options); 
    $select = $cnx->query("SELECT COUNT(*) as count FROM asterisk_cdr WHERE calldate LIKE '%2014-10-11%' AND channel LIKE '%SIP/4546975289%'"); 

    $select->setFetchMode(PDO::FETCH_OBJ); 

    while($row = $select->fetch()) 
    { 
     echo '<h1>', $row->count , '</h1>'; 
    } 
    } catch (Exception $e) { 
    echo "Connect failed : ", $e->getMessage(); 
    die(); 
    } 

Answer 2

你缺少一個引號（堆棧的語法高亮顯示了），但它應該有一個開口雙引號來代替，並用相同的結局。你不能使用所有的單引號。

我用一個雙引號取代了開頭的單引號，還有一個匹配的雙引號。

$mysql_query = mssql_query ("SELECT COUNT(*) FROM asterisk_cdr 
WHERE calldate LIKE '%2014-10-11%' 
    AND channel LIKE '%SIP/4546975289%'"); 

作爲旁註，您正在迴應錯誤的變量。

但是，這不是你如何迴應結果，而是一個循環。

喜歡的東西，並用一個取代Fieldname要使用：

while ($row = mssql_fetch_assoc($mysql_query)) { 

    print $row['Fieldname'] . "\n"; 

} 

或使用mssql_fetch_array()

您還可以使用：

$results= array(); 
while ($result = mssql_fetch_array($mysql_query)) { 
    $results[]= $result; 
} 

foreach($results as $result){ 
    echo $result['calldate'] . " " . $result['channel']; 
} 

---

欲瞭解更多信息在Microsoft SQL Server的功能上，請諮詢：

• http://php.net/manual/en/book.mssql.php

Answer 3

$mysql_query = mssql_query ('SELECT COUNT(*) FROM asterisk_cdr 
    WHERE calldate LIKE '%2014-10-11%' 
    AND channel LIKE '%SIP/4546975289%'); 

    while($row=mssql_fetch_array($mysql_query)) 
    { 
    echo $row[0]; 
    } 

Answer 4

$mysqli = new mysqli($servername, $username, $password, $dbname); 

// Check connection 
if ($mysqli->connect_error) { 
    die("Connection failed: " . $mysqli->connect_error); 
} 

$mysql_query = mysqli ("SELECT COUNT(*) FROM asterisk_cdr WHERE calldate LIKE '%2014-10-11%' AND channel LIKE '%SIP/4546975289%'"); 

while ($row = mysql_fetch_array($mysql_query, MYSQL_ASSOC)) { 
    echo ($row["channel"]); 
}