[英]Calling a mysql query inside another mysql query while loop
include "mysql.php";
$query= "SELECT ID,name,displayname,established,summary,searchlink,imagename,image FROM institutions ORDER BY rand() ";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo "<div class='grid-item item1' style='background: ".ran_col().";'>";
$query1= "SELECT content FROM ".$row['name']." limit 1";
$result1=mysql_query($query1,$db);
while($row1=mysql_fetch_array($result1))
{
echo "<div class='content-short' data-id=".$row['name'].">";
$string = $row1['content'];
if (strlen($string) > 200)
{
$trimstring = substr($string, 0,200). '...';
}
else
{
$trimstring = substr($string,0). '...';
}
echo $trimstring;
echo "</div>";
}
echo <"/div">;
}
應該執行什么代碼:
代碼應從機構表中獲取數據,然后將$row['name']
用作下一個mysql查詢的表,從該表中獲取$row['name']
表$row['content']
但是它不能像預期的那樣工作
您到處都使用相同的變量。 更改為
include "mysql.php";
$query= "SELECT ID,name,displayname,established,summary,searchlink,imagename,image FROM institutions ORDER BY rand() ";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo "<div class='grid-item item1' style='background: ".ran_col().";'>";
$query1= "SELECT content FROM ".$row['name']." limit 1";
$result1=mysql_query($query1,$db);
while($row1=mysql_fetch_array($result1))
{
echo "<div class='content-short' data-id=".$row['name'].">";
$string = $row1['content'];
if (strlen($string) > 200)
{
$trimstring = substr($string, 0,200). '...';
}
else
{
$trimstring = substr($string,0). '...';
}
echo $trimstring;
echo "</div>";
}
echo <"/div">;
}
編輯:
另外你的查詢是錯誤的
$query1= "SELECT content FROM ".$row['name']" limit 1";
至
$query1= "SELECT content FROM ".$row['name']." limit 1";
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