mysql - Parameter warning php -


इस सवाल का पहले से ही एक उत्तर है: < 33 उत्तरों 

  $ name1 = 
   $ _POST [ 'name1']; $ Lname = $ _POST ['lname']; $ Uname = $ _POST ['uname']; $ Email1 = $ email1; $ ईमेल 2 = $ ईमेल 2; $ Pass1 = $ pass1; $ Pass2 = $ pass2; $ पास 1 = एमडी 5 ($ पास 1); $ Sql ​​= "का चयन करें * उपयोगकर्ताओं से जहां uname = $ uname"; यदि (mysql_num_rows ($ sql) & gt; 0) {गूंज "क्षमा करें, वह उपयोगकर्ता पहले से मौजूद है"; बाहर जाएं(); }

चेतावनी: mysql_num_rows () पैरामीटर 1 को संसाधन होने की उम्मीद है, /nas/students/unix/public_html/dsa1/register.php में लाइन 27 पर दिया गया है

निश्चित रूप से:

int mysql_num_rows (संसाधन $ परिणाम)

पुनर्प्राप्त करता है एक परिणाम सेट से पंक्तियों की संख्या। यह कमांड केवल SELECT या SHOW जैसे एक बयान के लिए मान्य है जो कि वास्तविक परिणाम सेट को वापस लाती है।

एक संसाधन एक एसक्यूएल क्वेरी से कुछ अलग है इस मामले में एक संसाधन है जो PHP MySQL लाइब्रेरी एक कथन के संदर्भ में चारों ओर से गुजरता है जो पहले से ही चल रहा है।

आपको एसक्यूएल स्टेटमेंट को mysql_query को चलाने के लिए पहले चलाने की जरूरत है । mysql_num_rows तब उस फ़ंक्शन कॉल के नतीजे लेगा और आपको बताएगा कि उसमें कितनी पंक्तियाँ हैं।


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